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1ldprlg0x | maths | differential-equations | formation-of-differential-equations | <p> Let a differentiable function $$f$$ satisfy $$f(x)+\int_\limits{3}^{x} \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3$$. Then $$12 f(8)$$ is equal to :</p> | [{"identifier": "A", "content": "19"}, {"identifier": "B", "content": "34"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "1"}] | ["C"] | null | Differentiating both sides we get,
<br/><br/>$$
\begin{aligned}
& f^{\prime}(x)+\frac{f(x)}{x}=\frac{1}{2 \sqrt{x+1}} \\\\
& \Rightarrow \frac{d y}{d x}+\frac{y}{x}=\frac{1}{2 \sqrt{x+1}} \\\\
& \Rightarrow \mathrm{IF}=x \\\\
& \Rightarrow y x=\frac{1}{2} \int \frac{x}{\sqrt{x+1}} d x+c \\\\
& \Rightarrow y x=\frac{1}{2}\left(\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}-2(x+1)^{\frac{1}{2}}\right)+c \\\\
& x y=\frac{1}{3}(x+1)^{\frac{3}{2}}-(x+1)^{\frac{1}{2}}+c \\\\
& f(3)=2
\end{aligned}
$$
<br/><br/>So, $x=3, y=2$
<br/><br/>$\Rightarrow c=\frac{16}{3}$
<br/><br/>Now, $x=8$
<br/><br/>$8 f(8)=\frac{27}{3}-3+\frac{16}{3}=\frac{34}{3}$
<br/><br/>$12 f(8)=\frac{34}{3} \times \frac{12}{8}=17$ | mcq | jee-main-2023-online-31st-january-morning-shift |
lv5grwbn | maths | differential-equations | formation-of-differential-equations | <p>Let $$f(x)$$ be a positive function such that the area bounded by $$y=f(x), y=0$$ from $$x=0$$ to $$x=a>0$$ is $$e^{-a}+4 a^2+a-1$$. Then the differential equation, whose general solution is $$y=c_1 f(x)+c_2$$, where $$c_1$$ and $$c_2$$ are arbitrary constants, is</p> | [{"identifier": "A", "content": "$$\\left(8 e^x+1\\right) \\frac{d^2 y}{d x^2}-\\frac{d y}{d x}=0$$\n"}, {"identifier": "B", "content": "$$\\left(8 e^x+1\\right) \\frac{d^2 y}{d x^2}+\\frac{d y}{d x}=0$$\n"}, {"identifier": "C", "content": "$$\\left(8 e^x-1\\right) \\frac{d^2 y}{d x^2}-\\frac{d y}{d x}=0$$\n"}, {"identifier": "D", "content": "$$\\left(8 e^x-1\\right) \\frac{d^2 y}{d x^2}+\\frac{d y}{d x}=0$$"}] | ["B"] | null | <p>$$\int_\limits0^a f(x) d x=e^{-a}+4 a^2+a-1$$</p>
<p>Differentiate equation w.r.t. 'a'</p>
<p>$$\begin{aligned}
& f(a)=-e^{-a}+8 a+1 \\
& \Rightarrow f(x)=-e^{-x}+8 x+1
\end{aligned}$$</p>
<p>And $$y=c_1 f(x)+c_2$$</p>
<p>$$\begin{aligned}
& y=c_1\left(-e^{-x}+8 x+1\right)+c_2 \\
& y^{\prime}=c_1\left(e^{-x}+8\right) \Rightarrow c_1=\frac{y^{\prime}}{e^{-x}+8}
\end{aligned}$$</p>
<p>$$y^{\prime \prime}=-c_1 e^{-x}$$</p>
<p>put value of $$c_1$$</p>
<p>$$\begin{aligned}
& \frac{d^2 y}{d x^2}=\frac{-\frac{d y}{d x} \cdot e^{-x}}{\left(e^{-x}+8\right)}=\frac{\frac{d y}{d x}}{\left(1+8 e^x\right)} \\
& \Rightarrow\left(1+8 e^x\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=1
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
lv9s20hz | maths | differential-equations | formation-of-differential-equations | <p>The differential equation of the family of circles passing through the origin and having centre at the line $$y=x$$ is :</p> | [{"identifier": "A", "content": "$$\\left(x^2-y^2+2 x y\\right) \\mathrm{d} x=\\left(x^2-y^2+2 x y\\right) \\mathrm{d} y$$\n"}, {"identifier": "B", "content": "$$\\left(x^2+y^2-2 x y\\right) \\mathrm{d} x=\\left(x^2+y^2+2 x y\\right) \\mathrm{d} y$$\n"}, {"identifier": "C", "content": "$$\\left(x^2+y^2+2 x y\\right) \\mathrm{d} x=\\left(x^2+y^2-2 x y\\right) \\mathrm{d} y$$\n"}, {"identifier": "D", "content": "$$\\left(x^2-y^2+2 x y\\right) \\mathrm{d} x=\\left(x^2-y^2-2 x y\\right) \\mathrm{d} y$$"}] | ["D"] | null | <p>Equation of circle passing through origin & having centre at the line $$y=x$$ is</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwegsauf/ce0513c1-e394-4ced-87e0-f3bb34e8f595/28467680-1661-11ef-a209-0180f41f1307/file-1lwegsaug.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwegsauf/ce0513c1-e394-4ced-87e0-f3bb34e8f595/28467680-1661-11ef-a209-0180f41f1307/file-1lwegsaug.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Evening Shift Mathematics - Differential Equations Question 6 English Explanation"></p>
<p>$$\begin{aligned}
& (x-t)^2+(y-t)^2=2 t^2 \\
& x^2+y^2+t^2+t^2-2 t x-2 t y=2 t^2 \\
& x^2+y^2=2 t(x+y)
\end{aligned}$$</p>
<p>Now differentiate</p>
<p>$$\begin{aligned}
& 2 x+2 y y^{\prime}=2 t\left(1+y^{\prime}\right) \\
& t=\frac{x+y y^{\prime}}{1+y^{\prime}}
\end{aligned}$$</p>
<p>Now, $$x^2+y^2=2\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)(x+y)$$</p>
<p>$$\begin{aligned}
& x^2+y^2+x^2 \frac{d y}{d x}+y^2 \frac{d y}{d x} \\
& =2\left(x^2+x y+x y \frac{d y}{d x}+y^2 \frac{d y}{d x}\right) \\
& d x\left(x^2+y^2-2 x^2-2 x y\right)=d y\left(2 x y+2 y^2-x^2-y^2\right) \\
& d x\left(x^2-y^2+2 x y\right)=d y\left(x^2-y^2-2 x y\right)
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
40MGPqbhsQ7TtAf7 | maths | differential-equations | linear-differential-equations | The solution of the differential equation
<br/><br/>$$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,$$ is : | [{"identifier": "A", "content": "$$x{e^{2{{\\tan }^{ - 1}}y}} = {e^{{{\\tan }^{ - 1}}y}} + k$$ "}, {"identifier": "B", "content": "$$\\left( {x - 2} \\right) = k{e^{2{{\\tan }^{ - 1}}y}}$$ "}, {"identifier": "C", "content": "$$2x{e^{{{\\tan }^{ - 1}}y}} = {e^{2{{\\tan }^{ - 1}}y}} + k$$ "}, {"identifier": "D", "content": "$$x{e^{{{\\tan }^{ - 1}}y}} = {\\tan ^{ - 1}}y + k$$ "}] | ["C"] | null | $$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0$$
<br><br>$$ \Rightarrow {{dx} \over {dy}} + {x \over {\left( {1 + {y^2}} \right)}} = {{{e^{{{\tan }^{ - 1}}y}}} \over {\left( {1 + {y^2}} \right)}}$$
<br><br>$$I.F = e{}^{\int {{1 \over {\left( {1 + {y^2}} \right)}}dy} } = {e^{{{\tan }^{ - 1}}y}}$$
<br><br>$$x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = \int {{{{e^{{{\tan }^{ - 1}}y}}} \over {1 + {y^2}}}} {e^{{{\tan }^{ - 1}}y}}\,dy$$
<br><br>$$x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = {{{e^{2{{\tan }^{ - 1}}y}}} \over 2} + C$$
<br><br>$$\therefore$$ $$\,\,\,\,\,2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$$ | mcq | aieee-2003 |
66nkD6pi4hLZiTbu | maths | differential-equations | linear-differential-equations | The solution of the differential equation
<br/><br>$${{dy} \over {dx}} = {{x + y} \over x}$$ satisfying the condition $$y(1)=1$$ is :</br> | [{"identifier": "A", "content": "$$y = \\ln x + x$$ "}, {"identifier": "B", "content": "$$y = x\\ln x + {x^2}$$ "}, {"identifier": "C", "content": "$$y = x{e^{\\left( {x - 1} \\right)}}\\,$$ "}, {"identifier": "D", "content": "$$y = x\\,\\ln x + x$$ "}] | ["D"] | null | $${{dy} \over {dx}} = {{x + y} \over x} = 1 + {y \over x}$$
<br><br>Putting $$y=$$ $$vx$$
<br><br>and $${{dy} \over {dx}} = v + x{{dv} \over {dx}}$$
<br><br>we get
<br><br>$$v + x{{dv} \over {dx}} = 1 + v$$
<br><br>$$ \Rightarrow \int {{{dx} \over x}} = \int {dv} $$
<br><br>$$ \Rightarrow v = \ln {\mkern 1mu} x + c$$
<br><br>$$ \Rightarrow y = x\ln x + cx$$
<br><br>As $$\,\,\,\,y\left( 1 \right) = 1$$
<br><br>$$\therefore$$ $$c=1$$
<br><br>So solution is $$y=xlnx+x$$ | mcq | aieee-2008 |
MRSMdUCqIIT55v0x | maths | differential-equations | linear-differential-equations | If $${{dy} \over {dx}} = y + 3 > 0\,\,$$ and $$y(0)=2,$$ then $$y\left( {\ln 2} \right)$$ is equal to : | [{"identifier": "A", "content": "$$5$$"}, {"identifier": "B", "content": "$$13$$"}, {"identifier": "C", "content": "$$-2$$"}, {"identifier": "D", "content": "$$7$$"}] | ["D"] | null | $${{dy} \over {dx}} = y + 3 \Rightarrow \int {{{dy} \over {y + 3}}} = \int {dx} $$
<br><br>$$ \Rightarrow \ell n\left| {y + 3} \right| = x + c$$
<br><br>Since $$y\left( 0 \right) = 2,\,\,\,$$ $$\,\,\,\,\,\,\,$$ $$\therefore$$ $$\,\,\,\,\,\,\,$$ $$\ell n5 = c$$
<br><br>$$ \Rightarrow \ell n\left| {y + 3 = x + \ell n5} \right|$$
<br><br>When $$x = \ell n2,$$ then
<br><br>$$\ell n\left| {y + 3} \right| = \ell n2 + \ell n5$$
<br><br>$$ \Rightarrow \ln \left| {y + 3} \right| = \ell n\,10$$
<br><br>$$\therefore$$ $$\,\,\,\,\,$$ $$y + 3 = \pm 10 \Rightarrow y = 7, - 13$$ | mcq | aieee-2011 |
3IxYLRiMUbRQ20y2 | maths | differential-equations | linear-differential-equations | Let $$y(x)$$ be the solution of the differential equation
<br/><br>$$\left( {x\,\log x} \right){{dy} \over {dx}} + y = 2x\,\log x,\left( {x \ge 1} \right).$$ Then $$y(e)$$ is equal to :</br> | [{"identifier": "A", "content": "$$2$$ "}, {"identifier": "B", "content": "$$2e$$ "}, {"identifier": "C", "content": "$$e$$ "}, {"identifier": "D", "content": "$$0$$"}] | ["A"] | null | Given, $${{dy} \over {dx}} + \left( {{1 \over {x\,\log \,x}}} \right)y = 2$$
<br><br>$$I.F. = {e^{\int {{1 \over {x\log x}}dx} }} = {e^{\log \left( {\log x} \right)}} = \log x$$
<br><br>$$y.\log x = \int {2\,\log xdx + c} $$
<br><br>$$y\log x = 2\left[ {x\log x - x} \right] + c$$
<br><br>Put $$x=1,y.0=-2+c$$ $$ \Rightarrow c = 2$$
<br><br>Put $$x=e$$
<br><br>$$y\log e = 2e\left( {\log e - 1} \right) + c \Rightarrow y\left( e \right) = c = 2$$ | mcq | jee-main-2015-offline |
rnmxkZT2jG9mNT3hkCdZl | maths | differential-equations | linear-differential-equations | The solution of the differential equation
<br/><br/>$${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$$
<br/><br/>where 0 $$ \le $$ x < $${\pi \over 2}$$, and y (0) = 1, is given by : | [{"identifier": "A", "content": "y = 1 $$-$$ $${x \\over {\\sec x + \\tan x}}$$ "}, {"identifier": "B", "content": "y<sup>2</sup> = 1 + $${x \\over {\\sec x + \\tan x}}$$"}, {"identifier": "C", "content": "y<sup>2</sup> = 1 $$-$$ $${x \\over {\\sec x + \\tan x}}$$"}, {"identifier": "D", "content": "y = 1 + $${x \\over {\\sec x + \\tan x}}$$"}] | ["C"] | null | Given,
<br><br>$${{dy} \over {dx}} + {y \over 2}\sec x = {{\tan x} \over {2y}}$$
<br><br>$$ \Rightarrow $$ $$2y{{dy} \over {dx}} + {y^2}\sec x = \tan x$$
<br><br>Now, let
<br><br>y<sup>2</sup> $$=$$ t
<br><br>$$ \Rightarrow $$ 2y$${{dy} \over {dx}} = {{dt} \over {dx}}$$
<br><br>$$ \therefore $$ New equation,
<br><br>$${{dt} \over {dx}} + t\sec x = \tan x$$
<br><br>$$ \therefore $$ I.F $$=$$ $${e^{\int {\sec xdx} }}$$
<br><br>$$=$$ $${e^{\ln \left( {\sec x + \tan x} \right)}}$$
<br><br>$$=$$ sec x + tan x
<br><br>$$ \therefore $$ Solution is,
<br><br>t(sec x + tan x) $$=$$ $$\int {\tan x} $$ (sec x + tan x) dx
<br><br>$$ \Rightarrow $$ t(sec x + tan x) $$=$$ sec x + tan x $$-$$ x + c
<br><br>$$ \Rightarrow $$ t $$=$$ 1 $$-$$ $${x \over {\sec x + \tan x}} + c$$
<br><br>$$ \Rightarrow $$ y<sup>2</sup> $$=$$ 1 $$-$$ $${x \over {\sec x + \tan x}} + c$$
<br><br>Given,
<br><br>y(0) $$=$$ 1
<br><br>$$ \therefore $$ 1 $$=$$ 1 $$-$$ 0 + c
<br><br>$$ \Rightarrow $$ c $$=$$ 0
<br><br>$$ \therefore $$ y<sup>2</sup> $$=$$ 1 $$-$$ $${x \over {\sec x + \tan x}}$$ | mcq | jee-main-2016-online-10th-april-morning-slot |
3DimCCnBbdBvLDmU | maths | differential-equations | linear-differential-equations | If $$\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$$ and y(0) = 1,
<br/><br/>then $$y\left( {{\pi \over 2}} \right)$$ is equal to : | [{"identifier": "A", "content": "$$ - {2 \\over 3}$$"}, {"identifier": "B", "content": "$$ - {1 \\over 3}$$"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["D"] | null | $$\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$$
<br><br>$$ \Rightarrow $$ $${d \over {dx}}\left( {2 + \sin x} \right)\left( {y + 1} \right) = 0$$
<br><br>On integrating, we get
<br><br>(2 + sin x) (y + 1) = C
<br><br>At x = 0, y = 1 we have
<br><br>(2 + sin 0) (1 + 1) = C
<br><br>$$ \Rightarrow $$ C = 4
<br><br>$$ \Rightarrow $$ $$\left( {y + 1} \right) = {4 \over {2 + \sin x}}$$
<br><br>$$ \Rightarrow $$ y = $${4 \over {2 + \sin x}} - 1$$
<br><br>Now $$y\left( {{\pi \over 2}} \right) = {4 \over {2 + \sin {\pi \over 2}}} - 1$$
<br><br>= $${4 \over 3} - 1$$ = $${1 \over 3}$$ | mcq | jee-main-2017-offline |
GpKQ8v7At6kinAXrSFQ1D | maths | differential-equations | linear-differential-equations | The curve satisfying the differential equation, ydx $$-$$(x + 3y<sup>2</sup>)dy = 0 and passing through the point (1, 1), also passes through the point : | [{"identifier": "A", "content": "$$\\left( {{1 \\over 4}, - {1 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - {1 \\over 3},{1 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{1 \\over 3}, - {1 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 4}, {1 \\over 2}} \\right)$$"}] | ["B"] | null | Given,
<br><br>y dx = $$\left( {x + 3{y^2}} \right)dy$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ y $${{dx} \over {dy}}$$ = x + 3y<sup>2</sup>
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${{dx} \over {dy}}$$ $$-$$ $${x \over y} = 3y$$
<br><br>If = $${e^{ - \int {{1 \over y}dy} }}$$ = $${e^{ - \ln y}}$$ = $${1 \over y}$$
<br><br>$$\therefore\,\,\,$$ Soluation is ,
<br><br>x . $${1 \over y}$$ = $$\int {3y.{1 \over y}dy} $$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${x \over y}$$ = 3y + c
<br><br>This curve passing through (1, 1)
<br><br>$$\therefore\,\,\,$$ 1 = 3 + c
<br><br>$$ \Rightarrow $$$$\,\,\,$$ c = $$-$$ 2
<br><br>$$\therefore\,\,\,$$ Curve is, x = 3y<sup>2</sup> $$-$$ 2y
<br><br>Now put every point in this equation, and see which point satisfy this equation.
<br><br>Following this method you can see ($$-$$ $${1 \over 3}$$, $${1 \over 3}$$) point satisfy this equation. | mcq | jee-main-2017-online-8th-april-morning-slot |
Sj0uUY7qhaROkJ948JmxM | maths | differential-equations | linear-differential-equations | If 2x = y$${^{{1 \over 5}}}$$ + y$${^{ - {1 \over 5}}}$$ and
<br/><br/>(x<sup>2</sup> $$-$$ 1) $${{{d^2}y} \over {d{x^2}}}$$ + $$\lambda $$x $${{dy} \over {dx}}$$ + ky = 0,
<br/><br/>then $$\lambda $$ + k is equal to : | [{"identifier": "A", "content": "$$-$$ 23 "}, {"identifier": "B", "content": "$$-$$ 24 "}, {"identifier": "C", "content": "26 "}, {"identifier": "D", "content": "$$-$$ 26"}] | ["B"] | null | <p>It is given that</p>
<p>$$2x = {y^{1/5}} + {y^{ - 1/5}}$$</p>
<p>$$ \Rightarrow 2x = {y^{1/5}} + 1/{y^{1/5}}$$</p>
<p>Therefore, $$2x = a + {1 \over a} \Rightarrow {a^2} - 2ax + 1 = 0$$</p>
<p>$$a = {{2x \pm \sqrt {4{x^2} - 4} } \over 2}$$</p>
<p>$$ \Rightarrow a = {{2x \pm 2\sqrt {{x^2} - 1} } \over 2}$$</p>
<p>$$ \Rightarrow a = x \pm \sqrt {{x^2} - 1} $$</p>
<p>$$ \Rightarrow {y^{1/5}} = x \pm \sqrt {{x^2} - 1} $$</p>
<p>$$ \Rightarrow y = {(x \pm \sqrt {{x^2} - 1} )^5}$$</p>
<p>Therefore,</p>
<p>$${{dy} \over {dx}} = 5{(x \pm \sqrt {{x^2} - 1} )^4}\left( {1 \pm {{2x} \over {2\sqrt {{x^2} - 1} }}} \right)$$</p>
<p>$$ = 5{(x + \sqrt {{x^2} - 1} )^4}\left( {{{\sqrt {{x^2} - 1} \pm x} \over {\sqrt {{x^2} - 1} }}} \right)$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} = {{ - 5y} \over {\sqrt {{x^2} - 1} }}$$ ...... (1)</p>
<p>$$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {{\left[ {\sqrt {{x^2} - 1} \left( { - 5{{dy} \over {dx}}} \right) - 5( - 5y){1 \over 2}{{2x} \over {\sqrt {{x^2} - 1} }}} \right]} \over {({x^2} - 1)}}$$</p>
<p>Therefore, $$({x^2} - 1){{{d^2}y} \over {d{x^2}}} = - 5\sqrt {{x^2} - 1} {{dy} \over {dx}} + 5y{x \over {\sqrt {{x^2} - 1} }}$$</p>
<p>$$ \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 25y - x{{dy} \over {dx}}$$</p>
<p>$$ \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} + 1x{{dy} \over {dx}} - 25y = 0$$</p>
<p>Therefore, $$\lambda$$ = 1, k = $$-$$25; hence,</p>
<p>$$\lambda + k = - 24$$</p> | mcq | jee-main-2017-online-9th-april-morning-slot |
pTriHBV8FwZPwKpL | maths | differential-equations | linear-differential-equations | Let y = y(x) be the solution of the differential equation
<br/><br/>$$\sin x{{dy} \over {dx}} + y\cos x = 4x$$, $$x \in \left( {0,\pi } \right)$$.
<br/><br/>If $$y\left( {{\pi \over 2}} \right) = 0$$, then $$y\left( {{\pi \over 6}} \right)$$ is equal to : | [{"identifier": "A", "content": "$$ - {4 \\over 9}{\\pi ^2}$$"}, {"identifier": "B", "content": "$${4 \\over {9\\sqrt 3 }}{\\pi ^2}$$"}, {"identifier": "C", "content": "$$ - {8 \\over {9\\sqrt 3 }}{\\pi ^2}$$"}, {"identifier": "D", "content": "$$ - {8 \\over 9}{\\pi ^2}$$"}] | ["D"] | null | Given,
<br><br> sin x $${{dy} \over {dx}} + y\cos y = 4x$$
<br><br>$$ \Rightarrow \,\,\,\,{{dy} \over {dx}}\,$$ + y cot x = 4x cosec x
<br><br>This is a linear differential equation of form,
<br><br> $${{dy} \over {dx}}\,$$ + py = Q
<br><br>Where p = cot x and Q = 4x cosec x
<br><br>So, Integrating factor (I. F)
<br><br> = $${e^{\int {pdx} }}$$
<br><br>= $${e^{\int {\cot dx} }}$$
<br><br>= $${e^{\ln \left| {\sin \,x} \right|}}$$
<br><br= $$\left|="" {\sin="" x}="" \right|$$="" <br=""><br>= sin x as $$x \in \left( {0,\pi } \right)$$
<br><br>Solution of the differential equation is
<br><br>y sin x = $$\int {} $$ 4x cosecx sinx dx + c
<br><br>$$ \Rightarrow \,\,\,\,$$ y sinx = $$\int {4x\,dx\, + c} $$
<br><br>$$ \Rightarrow \,\,\,\,$$ y sinx = 4.$${{{x^2}} \over 2} + c$$
<br><br>$$ \Rightarrow \,\,\,\,$$ y sinx = 2x<sup>2</sup> + c . . . . . (1)
<br><br>Given that, $$y\left( {{\pi \over 2}} \right) = 0$$
<br><br>$$\therefore\,\,\,$$ x = $$y\left( {{\pi \over 2}} \right) = 0$$ and y = 0
<br><br>Put this x = $${\pi \over 2}$$ and y = 0 at equation (1)
<br><br>0.1 = 2. $$\left( {{\pi \over 2}} \right)$$<sup>2</sup> + c
<br><br>$$ \Rightarrow \,\,\,$$ c =$$ - {{{\pi ^2}} \over 2}$$
<br><br>So, differential equation is
<br><br>y sin x = 2x<sup>2</sup> $$-$$ $${{{\pi ^2}} \over 2}\,\,.....(2)$$
<br><br>Now we have to find y $$\left( {{\pi \over 6}} \right).$$
<br><br>So, put x = $${{\pi \over 6}}$$ at equation (2)
<br><br>y . sin $${{\pi \over 6}}$$ = 2 $${\left( {{\pi \over 6}} \right)^2} - {{{\pi ^2}} \over 2}$$
<br><br>$$ \Rightarrow \,\,\,\,\,\,\,y.{1 \over 2}$$ = 2. $${{{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 2}$$
<br><br>$$ \Rightarrow \,\,\,\,\,{y \over 2} = {{{\pi ^2}} \over {18}} - {{{\pi ^2}} \over 2}$$
<br><br>$$ \Rightarrow \,\,\,\,{y \over 2} = {{{\pi ^2} - 9{\pi ^2}} \over {18}}$$
<br><br>$$ \Rightarrow \,\,\,\,\,y = - {{8{\pi ^2}} \over 9}$$ </br=> | mcq | jee-main-2018-offline |
4oqPlOolimDvLDBtcZSne | maths | differential-equations | linear-differential-equations | If y = y(x) is the solution of the differential equation,
<br/><br/>x$$dy \over dx$$ + 2y = x<sup>2</sup>, satisfying y(1) = 1, then y($$1\over2$$) is equal
to : | [{"identifier": "A", "content": "$$ {{7} \\over {64}}$$"}, {"identifier": "B", "content": "$$ {{49} \\over {16}}$$"}, {"identifier": "C", "content": "$$ {{1} \\over {4}}$$"}, {"identifier": "D", "content": "$$ {{13} \\over {16}}$$"}] | ["B"] | null | Given,
<br><br>$$x{{dy} \over {dx}} + 2y = {x^2}$$
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} + \left( {{2 \over x}} \right)y = x$$
<br><br>This is a linear differential equation.
<br><br>$$ \therefore $$ I.F $$ = {e^{\int {{2 \over x}dx} }}$$
<br><br>$$ = {e^{2\ln x}}$$
<br><br>$$ = {x^2}$$
<br><br>$$ \therefore $$ Solution is,
<br><br>$$y \cdot {x^2} = \int {x \cdot {x^2}dx} $$
<br><br>$$ \Rightarrow $$ $$y{x^2} = {{{x^4}} \over 4} + C$$
<br><br>given $$y\left( 1 \right) = 1$$
<br><br>$$ \therefore $$ $$1.1 = {4 \over 4} + C$$
<br><br>$$ \Rightarrow $$ $$C = {3 \over 4}$$
<br><br>$$ \therefore $$ Equation is
<br><br>$$y{x^2} = {{{x^4}} \over 4} + {3 \over 4}$$
<br><br>$$ \therefore $$ $$y\left( {{1 \over 2}} \right)$$ means $$x = {1 \over 2}$$
<br><br>$$ \therefore $$ $$y \cdot {\left( {{1 \over 2}} \right)^2} = {1 \over 4} \times {\left( {{1 \over 2}} \right)^4} + {3 \over 4}$$
<br><br>$$ \Rightarrow $$ $${y \over 4} = {1 \over {64}} + {3 \over 4}$$
<br><br>$$ \Rightarrow $$ $${y \over 4} = {{1 + 48} \over {64}}$$
<br><br>$$ \Rightarrow $$ y = $${{49} \over {16}}$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
rgDzuaCRMrRjOzCFUmhTo | maths | differential-equations | linear-differential-equations | Let f be a differentiable function such that f '(x) = 7 - $${3 \over 4}{{f\left( x \right)} \over x},$$ (x > 0) and f(1) $$ \ne $$ 4. Then $$\mathop {\lim }\limits_{x \to 0'} \,$$ xf$$\left( {{1 \over x}} \right)$$ : | [{"identifier": "A", "content": "does not exist "}, {"identifier": "B", "content": "exists and equals $${4 \\over 7}$$"}, {"identifier": "C", "content": "exists and equals 4"}, {"identifier": "D", "content": "exists and equals 0"}] | ["C"] | null | $$f'(x) = 7 - {3 \over 4}{{f\left( x \right)} \over x}\,\,\,\left( {x > 0} \right)$$
<br><br>Given f(1) $$ \ne $$ 4 $$\mathop {\lim }\limits_{x \to {0^ + }} \,xf\left( {{1 \over x}} \right)\, = ?$$
<br><br>$${{dy} \over {dx}} + {3 \over 4}{y \over x} = 7$$ (This is LDE)
<br><br>IF $$ = {e^{\int {{3 \over {4x}}dx} }} = {e^{{3 \over 4}\ln \left| x \right|}} = {x^{{3 \over 4}}}$$
<br><br>$$y.{x^{{3 \over 4}}} = \int {7.{x^{{3 \over 4}}}} dx$$
<br><br>$$y.{x^{{3 \over 4}}} = 7.{{{x^{{7 \over 4}}}} \over {{7 \over 4}}} + C$$
<br><br>$$f(x) = 4x + C.{x^{ - {3 \over 4}}}$$
<br><br>$$f\left( {{1 \over 4}} \right) = {4 \over x} + C.{x^{{3 \over 4}}}$$
<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {4 + C.{x^{{7 \over 4}}}} \right) = 4$$ | mcq | jee-main-2019-online-10th-january-evening-slot |
SLAMDiu8qktuVsRJRGr1U | maths | differential-equations | linear-differential-equations | The solution of the differential equation,
<br/><br/>$${{dy} \over {dx}}$$ = (x – y)<sup>2</sup>, when y(1) = 1, is : | [{"identifier": "A", "content": "$$-$$ log<sub>e</sub> $$\\left| {{{1 + x - y} \\over {1 - x + y}}} \\right|$$ = x + y $$-$$ 2"}, {"identifier": "B", "content": "log<sub>e</sub> $$\\left| {{{2 - x} \\over {2 - y}}} \\right|$$ = x $$-$$ y"}, {"identifier": "C", "content": "log<sub>e</sub> $$\\left| {{{2 - y} \\over {2 - x}}} \\right|$$ = 2(y $$-$$ 1)"}, {"identifier": "D", "content": "$$-$$ log<sub>e</sub> $$\\left| {{{1 - x + y} \\over {1 + x - y}}} \\right|$$ = 2(x $$-$$ 1)"}] | ["D"] | null | x $$-$$ y = t <br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} = 1 - {{dt} \over {dx}}$$
<br><br>$$ \Rightarrow $$ 1 $$-$$ $${{dt} \over {dx}}$$ = t<sup>2</sup> $$ \Rightarrow $$ $$\int {{{dt} \over {1 - {t^2}}}} $$ = $$\int {1dx} $$
<br><br>$$ \Rightarrow $$ $${1 \over 2}\ell n\left( {{{1 + t} \over {1 - t}}} \right) = x + \lambda $$
<br><br>$$ \Rightarrow $$ $${1 \over 2}\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right) = x + \lambda $$ given y(1) = 1
<br><br>$$ \Rightarrow $$ $${1 \over 2}\ell n(1) = 1 + \lambda \Rightarrow \lambda = - 1$$
<br><br>$$ \Rightarrow $$ $$\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right)$$ = 2(x $$-$$ 1)
<br><br>$$ \Rightarrow $$ $$ - \ell n\left( {{{1 - x + y} \over {1 + x - y}}} \right)$$ = 2(x $$-$$ 1) | mcq | jee-main-2019-online-11th-january-evening-slot |
m4gBqYct3UWqiju5XSOgB | maths | differential-equations | linear-differential-equations | Let y = y(x) be the solution of the differential equation, x$${{dy} \over {dx}}$$ + y = x log<sub>e</sub> x, (x > 1). If 2y(2) = log<sub>e</sub> 4 $$-$$ 1, then y(e) is equal to : | [{"identifier": "A", "content": "$$ - {e \\over 2}$$"}, {"identifier": "B", "content": "$$ - {{{e^2}} \\over 2}$$"}, {"identifier": "C", "content": "$${{{e^2}} \\over 4}$$"}, {"identifier": "D", "content": "$${e \\over 4}$$"}] | ["D"] | null | $${{dy} \over {dx}} = {y \over x} = \ell nx$$
<br><br>$${e^{\int {{1 \over x}dx} }} = x$$
<br><br>$$xy = \int {x\ell nx + C} $$
<br><br>$$\ell nx{{{x^2}} \over 2} - \int {{1 \over x}.{{{x^2}} \over 2}} $$
<br><br>$$xy = {x \over 2}\ell nx - {{{x^2}} \over 4} + C,$$
<br><br>for $$2y\left( 2 \right) = 2\ell n2 - 1$$
<br><br>$$ \Rightarrow $$ $$C = 0$$
<br><br>$$y = {x \over 2}\ell nx - {x \over 4}$$
<br><br>$$y\left( e \right) = {e \over 4}$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
whv5C9v72fNRcGazLVCKP | maths | differential-equations | linear-differential-equations | If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as $${{{x^2} - 2y} \over x}$$, then the curve also passes through the point : | [{"identifier": "A", "content": "(\u20131, 2)"}, {"identifier": "B", "content": "$$\\left( { - \\sqrt 2 ,1} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { \\sqrt 3 ,0} \\right)$$"}, {"identifier": "D", "content": "(3, 0)"}] | ["C"] | null | $${{dy} \over {dx}} = {{{x^2} - 2y} \over x}$$ (Given)
<br><br>$${{dy} \over {dx}} + 2{y \over x} = x$$
<br><br>I.F = $${e^{\int {{2 \over x}dx} }} = {x^2}$$
<br><br>$$ \therefore $$ y.x<sup>2</sup> = $$\int {x.{x^2}} dx + C$$
<br><br>= $${{{x^4}} \over y} + C$$
<br><br>This curve passes through (1, $$-$$ 2) $$ \Rightarrow $$ C= $$-$$ $${9 \over 4}$$
<br><br>$$ \therefore $$ yx<sup>2</sup> = $${{{x^4}} \over 4} - {9 \over 4}$$
<br><br>Now check option(s), Which is satisly by option (ii) | mcq | jee-main-2019-online-12th-january-evening-slot |
gfALgYQ98fXlovzfMCNBL | maths | differential-equations | linear-differential-equations | The solution of the differential equation
<br/><br/>$$x{{dy} \over {dx}} + 2y$$ = x<sup>2</sup> (x $$ \ne $$ 0) with y(1) = 1, is : | [{"identifier": "A", "content": "$$y = {4 \\over 5}{x^3} + {1 \\over {5{x^2}}}$$"}, {"identifier": "B", "content": "$$y = {3 \\over 4}{x^2} + {1 \\over {4{x^2}}}$$"}, {"identifier": "C", "content": "$$y = {{{x^2}} \\over 4} + {3 \\over {4{x^2}}}$$"}, {"identifier": "D", "content": "$$y = {{{x^3}} \\over 5} + {1 \\over {5{x^2}}}$$"}] | ["C"] | null | $$x{{dy} \over {dx}} + 2y$$ = x<sup>2</sup>
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} + \left( {{2 \over x}} \right)y = x$$
<br><br>$$ \therefore $$ I.F = $${e^{\int {{2 \over x}dx} }}$$ = x<sup>2</sup>
<br><br>$$ \therefore $$ The solution is
<br><br>yx<sup>2</sup> = $$\int {{x^3}dx} $$
<br><br>$$ \Rightarrow $$ yx<sup>2</sup> = $${{{x^4}} \over 4} + C$$ .....(1)
<br><br>As y(1) = 1
<br><br>$$ \therefore $$ when x = 1 then y = 1.
<br><br>Putting the value of x and y in equation (1), we get
<br><br> 1 = $${1 \over 4} + C$$
<br><br>$$ \Rightarrow $$ C = $${3 \over 4}$$
<br><br>$$ \therefore $$ Required solution
<br><br>yx<sup>2</sup> = $${{{x^4}} \over 4} + {3 \over 4}$$
<br><br>$$ \Rightarrow $$ $$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$$ | mcq | jee-main-2019-online-9th-april-morning-slot |
99dcQh1k3EskUV7L8m18hoxe66ijvwuqtuw | maths | differential-equations | linear-differential-equations | If $$\cos x{{dy} \over {dx}} - y\sin x = 6x$$, (0 < x < $${\pi \over 2}$$)<br/>
and $$y\left( {{\pi \over 3}} \right)$$ = 0 then $$y\left( {{\pi \over 6}} \right)$$ is equal to :- | [{"identifier": "A", "content": "$$ - {{{\\pi ^2}} \\over {2 }}$$"}, {"identifier": "B", "content": "$$ - {{{\\pi ^2}} \\over {4\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$$ {{{\\pi ^2}} \\over {2\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$$ - {{{\\pi ^2}} \\over {2\\sqrt 3 }}$$"}] | ["D"] | null | $$\cos x{{dy} \over {dx}} - y\sin x = 6x$$
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} - y\tan x = 6x.\sec x$$
<br><br>This is a linear differential equation.
<br><br>$$ \therefore $$ I.F = $${e^{\int { - \tan xdx} }}$$ = $$\left| {\cos x} \right|$$
<br><br>As 0 < x < $${\pi \over 2}$$, so I.F = $$\cos x$$
<br><br>So solution is
<br><br>y(I.F) = $$\int {6x.\sec x\left( {I.F} \right)} dx$$
<br><br>$$ \Rightarrow $$ y cosx = $$\int {6x.\sec x\cos x} dx$$
<br><br>$$ \Rightarrow $$ y cosx = $$6\int x dx$$
<br><br>$$ \Rightarrow $$ y cosx = 3x<sup>2</sup> + C
<br><br>As $$y\left( {{\pi \over 3}} \right) = 0 $$<br><br>
$$ \therefore $$ $$ \left( 0 \right) \times \left( {{1 \over 2}} \right) = {{3{\pi ^2}} \over 9} + C $$
<br><br>$$\Rightarrow C = {{ - {\pi ^2}} \over 3}$$<br><br>
$$ \Rightarrow y\cos x = 3{x^2} - {{{\pi ^2}} \over 3}$$<br><br>
For $$y\left( {{\pi \over 6}} \right)$$<br><br>
$$y{{\sqrt 3 } \over 2} = {{3{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 3}$$<br><br>
$$ \Rightarrow $$ $${{\sqrt 3 y} \over 2} = {{ - 3{\pi ^2}} \over {12}} $$
<br><br>$$\Rightarrow y = {{ - {\pi ^2}} \over {2\sqrt 3 }}$$ | mcq | jee-main-2019-online-9th-april-evening-slot |
6tKArwY0tvD0aNwYof3rsa0w2w9jwxjytgt | maths | differential-equations | linear-differential-equations | If y = y(x) is the solution of the differential equation
<br/>$${{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$,
<br/>such that y (0) = 0, then $$y\left( { - {\pi \over 4}} \right)$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 2} - e$$"}, {"identifier": "B", "content": "$$e - 2$$"}, {"identifier": "C", "content": "$$2 + {1 \\over e}$$"}, {"identifier": "D", "content": "$${1 \\over e} - 2$$"}] | ["B"] | null | $${{dy} \over {dx}} + y{\sec ^2}x = se{c^2}x\,tanx$$<br> $$ \to $$ This is linear
differential equation<br><br>
$$IF = {e^{\int {{{\sec }^2}xdx} }} = {e^{\tan x}}$$<br><br>
Now solution is<br><br>
$$y.{e^{\tan x}} = \int {{e^{\tan x}}} {\sec ^2}x\tan xdx$$<br><br>
$$ \therefore $$ Let tanx = t<br><br>
sec<sup>2</sup>xdx = dt<br><br>
$$y.{e^{\tan x}} = \int {\mathop {{e^t}}\limits_{||} } \mathop t\limits_| dt$$<br><br>
ye<sup>tanx</sup> = te<sup>t</sup> – e<sup>t</sup> + c<br><br>
ye<sup>tanx</sup> = (tanx – 1)e<sup>tanx</sup> + c<br><br>
y = (tanx – 1) + c · e<sup>–tanx</sup><br><br>
Given y(0) = 0
<br><br>
$$ \Rightarrow $$ 0 = –1 + c $$ \Rightarrow $$ c = 1<br><br>
$$y\left( { - {\pi \over 4}} \right) = - 1 - 1 + e = - 2 + e$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
WM9oDlLJMOCMmXalVp3rsa0w2w9jx2e44h5 | maths | differential-equations | linear-differential-equations | Let y = y(x) be the solution of the differential equation,
<br/>$${{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, such that
y(0) = 1. Then : | [{"identifier": "A", "content": "$$y\\left( {{\\pi \\over 4}} \\right) - y\\left( { - {\\pi \\over 4}} \\right) = \\sqrt 2 $$"}, {"identifier": "B", "content": "$$y'\\left( {{\\pi \\over 4}} \\right) - y'\\left( { - {\\pi \\over 4}} \\right) = \\pi - \\sqrt 2 $$"}, {"identifier": "C", "content": "$$y\\left( {{\\pi \\over 4}} \\right) + y\\left( { - {\\pi \\over 4}} \\right) = {{{\\pi ^2}} \\over 2} + 2$$"}, {"identifier": "D", "content": "$$y'\\left( {{\\pi \\over 4}} \\right) + y'\\left( { - {\\pi \\over 4}} \\right) = - \\sqrt 2 $$"}] | ["B"] | null | $${{dy} \over {dx}} + y(\tan x) = 2x + {x^2}\tan x$$<br><br>
I.F. = $${e^{\int {\tan x\,dx} }}$$ = $${e^{\ln \sec x}}$$ = sec x<br><br>
y. sec x = $$\int {(2x + {x^2}\tan x)\sec \,x\,dx} $$<br><br>
$$ \Rightarrow y\sec x = {x^2}\sec x + \lambda $$<br><br>
$$ \Rightarrow $$ y(0) = 0 + $$\lambda $$ = 1 $$ \Rightarrow $$ $$\lambda $$ = 1<br><br>
$$ \Rightarrow $$ y = x<sup>2</sup> + cos x<br><br>
$$ \Rightarrow $$ y' = 2x – sinx<br><br>
$$ \Rightarrow y'\left( {{\pi \over 4}} \right) = {\pi \over 2} - {1 \over {\sqrt 2 }}$$<br><br>
$$ \Rightarrow y'\left( { - {\pi \over 4}} \right) = - {\pi \over 2} + {1 \over {\sqrt 2 }}$$<br><br>
$$ \therefore $$ $$y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2 $$ | mcq | jee-main-2019-online-10th-april-evening-slot |
bnqqrySjL2UYF3KIS73rsa0w2w9jx61ln0t | maths | differential-equations | linear-differential-equations | Consider the differential equation, $${y^2}dx + \left( {x - {1 \over y}} \right)dy = 0$$, If value of y is 1 when x = 1, then the value of x
for which y = 2, is : | [{"identifier": "A", "content": "$${3 \\over 2} - {1 \\over {\\sqrt e }}$$"}, {"identifier": "B", "content": "$${1 \\over 2} + {1 \\over {\\sqrt e }}$$"}, {"identifier": "C", "content": "$${5 \\over 2} + {1 \\over {\\sqrt e }}$$"}, {"identifier": "D", "content": "$${3 \\over 2} - \\sqrt e $$"}] | ["A"] | null | $${y^2}dx + \left( {x - {1 \over y}} \right)dy = 0$$<br><br>
$$ \Rightarrow {{dx} \over {dy}} + {x \over {{y^2}}} = {1 \over {{y^3}}}$$<br><br>
Integrating factor (I.F) = $${e^{ - {1 \over y}}}$$<br><br>
Now x.$${e^{ - {1 \over y}}}$$ = $$\int {{e^{ - {1 \over y}}}{1 \over {{y^3}}}dy} $$ by putting $${ - {1 \over y}}$$ = t<br><br>
x.e<sup>t</sup> = $$\int {{e^t}( - t)dt} $$<br><br>
$$ \Rightarrow x{e^t} = - (t.{e^t} - {e^t}) + c$$<br><br>
$$ \Rightarrow x{e^{ - {1 \over y}}} = {e^{ - {1 \over y}}}\left( {1 + {1 \over y}} \right) + c$$<br><br>
$$ \Rightarrow x = 1 + {1 \over y} + c.e^{1 \over y}$$<br><br>
It passes through (1, 1)<br><br>
$$ \therefore $$ c = $$-{1 \over e}$$<br><br>
Equation of the curve is <br><br>
$$x = 1 + {1 \over y} - {e^{{1 \over y} - 1}}$$
It passes through (k, 2)<br><br>
$$ \therefore $$ k = $$1 + {1 \over 2} - {e^{{1 \over 2} - 1}}$$
= $$3 \over 2$$ - $$1 \over{\sqrt e}$$
| mcq | jee-main-2019-online-12th-april-morning-slot |
F9aA9i4QzNhF9tJ7FRjgy2xukewqnyxo | maths | differential-equations | linear-differential-equations | Let y = y(x) be the solution of the differential
equation,
<br/>$${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$$, y > 0,y(0) = 1.<br/> If y($$\pi $$) = a and $${{dy} \over {dx}}$$ at x =
$$\pi $$ is b, then the ordered pair
(a, b) is equal to : | [{"identifier": "A", "content": "(2, 1)"}, {"identifier": "B", "content": "$$\\left( {2,{3 \\over 2}} \\right)$$"}, {"identifier": "C", "content": "(1, -1)"}, {"identifier": "D", "content": "(1, 1)"}] | ["D"] | null | $${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$$
<br><br>$$ \Rightarrow $$ $$\int {{{dy} \over {y + 1}}} = \int {{{\left( { - \cos x} \right)dx} \over {2 + \sin x}}} $$
<br><br>$$ \Rightarrow $$ $$\ln \left| {y + 1} \right| = - \ln \left| {2 + \sin x} \right| + \ln c$$
<br><br>$$ \Rightarrow $$ $$\ln \left| {\left( {y + 1} \right)\left( {2 + \sin x} \right)} \right| = \ln c$$
<br><br>As y(0) = 1
<br><br>$$ \therefore $$ $$\ln \left| {\left( {1 + 1} \right)\left( {2 + 0} \right)} \right| = \ln c$$
<br><br>$$ \Rightarrow $$ $$ \therefore $$ c = 4
<br><br>$$ \therefore $$ $${\left( {y + 1} \right)\left( {2 + \sin x} \right)}$$ = 4
<br><br>$$ \Rightarrow $$ y = $${{{2 - \sin x} \over {2 + \sin x}}}$$
<br><br>Given y($$\pi $$) = a
<br><br>$$ \therefore $$ y($$\pi $$) = $${{{2 - \sin \pi } \over {2 + \sin \pi }}}$$ = 1 = a
<br><br>$${{dy} \over {dx}} = {{\left( {2 + \sin x} \right)\left( { - \cos x} \right) - \left( {2 - \sin x} \right).\left( {\cos x} \right)} \over {{{\left( {2 + \sin x} \right)}^2}}}$$
<br><br>$$ \Rightarrow $$ $${\left. {{{dy} \over {dx}}} \right|_{x = \pi }} = $$ 1 = b
<br><br>So, (a, b) = (1, 1) | mcq | jee-main-2020-online-2nd-september-morning-slot |
lY7025GtrZtfgaBgayjgy2xukfqbmwrl | maths | differential-equations | linear-differential-equations | Let y = y(x) be the solution of the differential
equation
<br/><br>cosx$${{dy} \over {dx}}$$ + 2ysinx = sin2x, x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$.
<br/><br>If
y$$\left( {{\pi \over 3}} \right)$$ = 0, then y$$\left( {{\pi \over 4}} \right)$$ is equal to :</br></br> | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }} - 1$$"}, {"identifier": "B", "content": "$${\\sqrt 2 - 2}$$"}, {"identifier": "C", "content": "$${2 - \\sqrt 2 }$$"}, {"identifier": "D", "content": "$${2 + \\sqrt 2 }$$"}] | ["B"] | null | cosx$${{dy} \over {dx}}$$ + 2ysinx = sin2x
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} + 2y\tan x = 2\sin x$$
<br><br>I.F = $${e^{\int {2\tan xdx} }}$$ = sec<sup>2</sup> x
<br><br>y.sec<sup>2</sup> x = $${\int {2\sin x{{\sec }^2}xdx} }$$
<br><br>$$ \Rightarrow $$ ysec<sup>2</sup>x = $${\int {2\tan x\sec xdx} }$$
<br><br>$$ \Rightarrow $$ ysec<sup>2</sup>x = 2secx + c
<br><br>Given at x = $${\pi \over 3}$$, y = 0
<br><br>$$ \Rightarrow $$ 0 = $$2\sec {\pi \over 3} + c$$
<br><br>$$ \Rightarrow $$ c = -4
<br><br>ysec<sup>2</sup>x = 2secx - 4
<br><br>Here put x = $${\pi \over 4}$$
<br><br>y.2 = $$2\sqrt 2 $$ - 4
<br><br>$$ \Rightarrow $$ y = $$\sqrt 2 - 2$$ | mcq | jee-main-2020-online-5th-september-evening-slot |
bKLPl8x59ic5sk0Yvo1kls4nhfs | maths | differential-equations | linear-differential-equations | If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is $${{{x^2} - 4x + y + 8} \over {x - 2}}$$, then this curve also passes through the point : | [{"identifier": "A", "content": "(4, 4)"}, {"identifier": "B", "content": "(5, 5)"}, {"identifier": "C", "content": "(5, 4)"}, {"identifier": "D", "content": "(4, 5)"}] | ["B"] | null | Given<br><br>y (0) = 0<br><br>& $${{dy} \over {dx}} = {{{{(x - 2)}^2} + y + 4} \over {x - 2}}$$<br><br>$$ \Rightarrow {{dy} \over {dx}} - {y \over {x - 2}} = (x - 2) + {4 \over {x - 2}}$$<br><br>$$ \Rightarrow I.F. = {e^{ - \int {{1 \over {x - 2}}dx} }} = {1 \over {x - 2}}$$<br><br>Solution of D.E.<br><br>$$ \Rightarrow y.{1 \over {x - 2}} = \int {{1 \over {x - 2}}\left( {(x - 2) + {4 \over {x - 2}}} \right)} \,.\,dx$$<br><br>$$ \Rightarrow {y \over {x - 2}} = x - {4 \over {x - 2}} + C$$<br><br>Now, at x = 0, y = 0 $$ \Rightarrow $$ C = $$-$$2<br><br>$$ \therefore $$ y = x (x $$-$$ 2) $$-$$ 4 $$-$$ 2 (x $$-$$ 2)<br><br>$$ \Rightarrow $$ y = x<sup>2</sup> $$-$$ 4x<br><br>This curve passes through (5, 5) | mcq | jee-main-2021-online-25th-february-morning-slot |
wCQAbEV5LRYfKdBkJK1klta5pz7 | maths | differential-equations | linear-differential-equations | If the curve, y = y(x) represented by the solution of the differential equation (2xy<sup>2</sup> $$-$$ y)dx + xdy = 0, passes through the intersection of the lines, 2x $$-$$ 3y = 1 and 3x + 2y = 8, then |y(1)| is equal to _________. | [] | null | 1 | Given, <br><br>$$(2x{y^2} - y)dx + xdx = 0$$<br><br>$$ \Rightarrow {{dy} \over {dx}} + 2{y^2} - {y \over x} = 0$$<br><br>$$ \Rightarrow - {1 \over {{y^2}}}{{dy} \over {dx}} + {1 \over y}\left( {{1 \over x}} \right) = 2$$<br><br>$${1 \over y} = z$$<br><br>$$ - {1 \over {{y^2}}}{{dy} \over {dx}} = {{dz} \over {dx}}$$<br><br>$$ \Rightarrow {{dz} \over {dx}} + z\left( {{1 \over x}} \right) = 2$$<br><br>I. F. $$ = {e^{\int {{1 \over x}dx} }} = x$$<br><br>$$ \therefore $$ $$z(x) = \int {2(x)dx} = {x^2} + c$$<br><br>$$ \Rightarrow {x \over y} = {x^2} + c$$<br><br>As it passes through P(2, 1)<br><br>[Point of intersection of $$2x - 3y = 1$$ and $$3x + 2y = 8$$]<br><br>$$ \therefore $$ $${2 \over 1} = 4 + c$$<br><br>$$ \Rightarrow c = - 2$$<br><br>$$ \Rightarrow {x \over y} = {x^2} - 2$$<br><br>Put x = 1<br><br>$${1 \over y} = 1 - 2 = - 1$$<br><br>$$ \Rightarrow y(1) = - 1$$<br><br>$$ \Rightarrow |y(1)|\, = 1$$ | integer | jee-main-2021-online-25th-february-evening-slot |
pUtaxcN4gjjvU9lCae1kluhugkr | maths | differential-equations | linear-differential-equations | If y = y(x) is the solution of the equation <br/><br>$${e^{\sin y}}\cos y{{dy} \over {dx}} + {e^{\sin y}}\cos x = \cos x$$, y(0) = 0; then <br/><br>$$1 + y\left( {{\pi \over 6}} \right) + {{\sqrt 3 } \over 2}y\left( {{\pi \over 3}} \right) + {1 \over {\sqrt 2 }}y\left( {{\pi \over 4}} \right)$$ is equal to ____________.</br></br> | [] | null | 1 | e<sup>sin y</sup> cos y$${{dy} \over {dx}}$$ + e<sup>sin y</sup> cos x = cos x<br><br>Put e<sup>sin y</sup> = t<br><br>e<sup>sin y</sup> $$\times$$ cos y$${{dy} \over {dx}}$$ = $${{dt} \over {dx}}$$<br><br>$$ \Rightarrow $$ $${{dt} \over {dx}}$$ + t cos x = cos x<br><br>I. F. = $${e^{\int {\cos x\,dx} }} = {e^{\sin x}}$$<br><br>Solution of differential equation :<br><br>$$t.{e^{\sin x}} = \int {{e^{\sin x}}.\cos x\,dx} $$<br><br>$${e^{\sin y}}.{e^{\sin x}} = {e^{\sin x}} + c$$<br><br>at x = 0, y = 0<br><br>1 = 1 + c $$ \Rightarrow $$ c = 0<br><br>$$ \therefore $$ e<sup>sin x + sin y</sup> = e<sup>sin x</sup><br><br>$$ \Rightarrow $$ sin x + sin y = sin x<br><br>$$ \Rightarrow $$ sin y = 0 $$ \Rightarrow $$ y = 0<br><br>$$ \Rightarrow y\left( {{\pi \over 6}} \right) = 0,y\left( {{\pi \over 3}} \right) = 0,y\left( {{\pi \over 4}} \right) = 0$$
<br><br>$$ \therefore $$ $$1 + y\left( {{\pi \over 6}} \right) + {{\sqrt 3 } \over 2}y\left( {{\pi \over 3}} \right) + {1 \over {\sqrt 2 }}y\left( {{\pi \over 4}} \right)$$
<br><br>= 1 + 0 + 0 + 0 = 1 | integer | jee-main-2021-online-26th-february-morning-slot |
14bMOFYlViUBQmzhHc1kmhz8mfm | maths | differential-equations | linear-differential-equations | If y = y(x) is the solution of the differential equation, <br/><br/>$${{dy} \over {dx}} + 2y\tan x = \sin x,y\left( {{\pi \over 3}} \right) = 0$$, then the maximum value of the function y(x) over R is equal to: | [{"identifier": "A", "content": "$${1 \\over 8}$$"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "$$-$$$${15 \\over 4}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["A"] | null | $${{dy} \over {dx}} + 2\tan x.y = \sin x$$<br><br>$$I.F. = {e^{2\ln (\sec x)}} = {\sec ^2}x$$<br><br>$$y.{\sec ^2}x = \int {\sin x{{\sec }^2}xdx = \int {\tan x\sec x\,dx + c} } $$<br><br>$$y{\sec ^2}x = \sec x + c$$<br><br>$$y = \cos x + c{\cos ^2}x$$<br><br>$$x = {\pi \over 3},y = 0$$<br><br>$$ \Rightarrow {1 \over 2} + {c \over 4} \Rightarrow c = - 2$$<br><br>$$ \therefore $$ $$y = \cos x - 2{\cos ^2}x$$<br><br>$$y = - 2\left( {{{\cos }^2}x - {1 \over 2}\cos x} \right) = - 2\left( {{{\left( {\cos x - {1 \over 4}} \right)}^2} - {1 \over {16}}} \right)$$<br><br>$$y = {1 \over 8} - 2{\left( {\cos x - {1 \over 4}} \right)^2}$$<br><br>$$ \therefore $$ $${y_{\max }} = {1 \over 8}$$ | mcq | jee-main-2021-online-16th-march-morning-shift |
kpDSmbBYYMbv7EOUwD1kmivvbvq | maths | differential-equations | linear-differential-equations | If y = y(x) is the solution of the differential equation <br/><br/>$${{dy} \over {dx}}$$ + (tan x) y = sin x, $$0 \le x \le {\pi \over 3}$$, with y(0) = 0, then $$y\left( {{\pi \over 4}} \right)$$ equal to : | [{"identifier": "A", "content": "$${1 \\over 2}$$log<sub>e</sub> 2"}, {"identifier": "B", "content": "$$\\left( {{1 \\over {2\\sqrt 2 }}} \\right)$$ log<sub>e</sub> 2"}, {"identifier": "C", "content": "log<sub>e</sub> 2"}, {"identifier": "D", "content": "$${1 \\over 4}$$ log<sub>e</sub> 2"}] | ["B"] | null | Integrating Factor $$= {e^{\int {\tan x\,dx} }} = {e^{\ln (\sec x)}} = \sec x$$<br><br>$$y\sec x = \int {(\sin x)\sec x\,dx = \ln (\sec x) + C} $$<br><br>$$y(0) = 0 \Rightarrow C = 0$$<br><br>$$ \therefore $$ $$y = \cos x\ln |\sec x|$$<br><br>$$y\left( {{\pi \over 4}} \right) = {1 \over {\sqrt 2 }}\ln \left( {\sqrt 2 } \right) = {1 \over {2\sqrt 2 }}\ln 2$$ | mcq | jee-main-2021-online-16th-march-evening-shift |
zgBCh2waK1nNexvl3j1kmkmayl5 | maths | differential-equations | linear-differential-equations | If the curve y = y(x) is the solution of the differential equation<br/><br/> $$2({x^2} + {x^{5/4}})dy - y(x + {x^{1/4}})dx = {2x^{9/4}}dx$$, x > 0 which <br/><br/>passes through the point $$\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)$$, then the value of y(16) is equal to : | [{"identifier": "A", "content": "$$4\\left( {{{31} \\over 3} - {8 \\over 3}{{\\log }_e}3} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{{31} \\over 3} - {8 \\over 3}{{\\log }_e}3} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{{31} \\over 3} + {8 \\over 3}{{\\log }_e}3} \\right)$$"}, {"identifier": "D", "content": "$$4\\left( {{{31} \\over 3} + {8 \\over 3}{{\\log }_e}3} \\right)$$"}] | ["A"] | null | $${{dy} \over {dx}} - {y \over {2x}} = {{{x^{9/4}}} \over {{x^{5/4}}({x^{3/4}} + 1)}}$$<br><br>$$IF = {e^{ - \int {{{dx} \over {2x}}} }} = {e^{ - {1 \over 2}\ln x}} = {1 \over {{x^{1/2}}}}$$<br><br>$$y.{x^{ - 1/2}} = \int {{{{x^{9/4}}.{x^{ - 1/2}}} \over {{x^{5/4}}({x^{3/4}} + 1)}}} dx$$<br><br>= $$\int {{{{x^{1/2}}} \over {({x^{3/4}} + 1)}}} dx$$<br><br>Let, $$x = {t^4} \Rightarrow dx = 4{t^3}dt$$<br><br>= $$\int {{{{t^2}.4{t^3}dt} \over {({t^3} + 1)}}} $$<br><br>= $$4\int {{{{t^2}({t^3} + 1 - 1)} \over {({t^3} + 1)}}} d$$<br><br>= $$4\int {{t^2}dt - 4\int {{{{t^2}} \over {{t^3} + 1}}dt} } $$<br><br>= $${{4{t^3}} \over 3} - {4 \over 3}\ln ({t^3} + 1) + C$$<br><br>$$y{x^{ - 1/2}} = {{4{x^{3/4}}} \over 3} - {4 \over 3}\ln ({x^{3/4}} + 1) + C$$
<br><br>It passes through the point $$\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)$$
<br><br>$$ \therefore $$ $$1 - {4 \over 3}{\log _e}2 = {4 \over 3} - {4 \over 3}{\log _e}2 + C$$<br><br>$$ \Rightarrow C = - {1 \over 3}$$<br><br>$$y = {4 \over 3}{x^{5/4}} - {4 \over 3}\sqrt x \ln ({x^{3/4}} + 1) - {{\sqrt x } \over 3}$$<br><br>$$y(16) = {4 \over 3} \times 32 - {4 \over 3} \times 4\ln 9 - {4 \over 3}$$<br><br>$$ = {{124} \over 3} - {{32} \over 3}\ln 3 = 4\left( {{{31} \over 3} - {8 \over 3}\ln 3} \right)$$ | mcq | jee-main-2021-online-17th-march-evening-shift |
lFggJKW2JqxolmUfMm1kmkni3ai | maths | differential-equations | linear-differential-equations | Let y = y(x) be the solution of the differential equation <br/><br/>$$\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx,0 \le x \le {\pi \over 2},y(0) = 0$$. Then, $$y\left( {{\pi \over 3}} \right)$$ is equal to : | [{"identifier": "A", "content": "$$2{\\log _e}\\left( {{{\\sqrt 3 + 7} \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$2{\\log _e}\\left( {{{3\\sqrt 3 - 8} \\over 4}} \\right)$$"}, {"identifier": "C", "content": "$$2{\\log _e}\\left( {{{2\\sqrt 3 + 10} \\over {11}}} \\right)$$"}, {"identifier": "D", "content": "$$2{\\log _e}\\left( {{{2\\sqrt 3 + 9} \\over 6}} \\right)$$"}] | ["C"] | null | $$\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx$$ ..... (1)<br><br>$$(3\sin x + \cos x + 3)(\cos x\,dy - y\sin x\,dx) = dx$$<br><br>$$\int {d(y.\cos x) = \int {{{dx} \over {3\sin x + \cos x + 3}}} } $$<br><br>$$y\cos x = \int {{1 \over {3\left( {{{2 + \tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \right) + \left( {{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \right) + 3}}} $$<br><br>$$y\cos x = \int {{{{{\sec }^2}{x \over 2}} \over {6\tan {x \over 2} + 1 - {{\tan }^2}{x \over 2} + 3 + 3{{\tan }^2}{x \over 2}}}} $$<br><br>$$y\cos x = \int {{{{{\sec }^2}{x \over 2}} \over {2{{\tan }^2}{x \over 2} + 6\tan {x \over 2} + 4}}} = \int {{{{1 \over 2}{{\sec }^2}{x \over 2}dx} \over {{{\tan }^2}{x \over 2} + 3\tan {x \over 2} + 2}}} $$<br><br>$$y\cos x = \ln \left| {{{\tan {x \over 2} + 1} \over {\tan {x \over 2} + 2}}} \right| + c$$<br><br>Put x = 0 & y = 0<br><br>$$C = - \ln \left( {{1 \over 2}} \right) = \ln (2)$$<br><br>$$y\left( {{\pi \over 3}} \right) = 2\ln \left| {{{1 + \sqrt 3 } \over {1 + 2\sqrt 3 }}} \right| + \ln 2$$<br><br>$$ = 2\ln \left| {{{5 + \sqrt 3 } \over {11}}} \right| + \ln 2$$<br><br>$$ = 2\ln \left| {{{2\sqrt 3 + 10} \over {11}}} \right|$$ | mcq | jee-main-2021-online-17th-march-evening-shift |
OYJBP5zm1ymXBECXsW1kmm2hyx0 | maths | differential-equations | linear-differential-equations | Let y = y(x) be the solution of the differential equation <br/><br/>$${{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{x^2}/2}} - x} \right)$$, 0 < x < 2.1, with y(2) = 0. Then the value of $${{dy} \over {dx}}$$ at x = 1 is equal to : | [{"identifier": "A", "content": "$${{{e^{5/2}}} \\over {{{(1 + {e^2})}^2}}}$$"}, {"identifier": "B", "content": "$${{5{e^{1/2}}} \\over {{{({e^2} + 1)}^2}}}$$"}, {"identifier": "C", "content": "$$ - {{2{e^2}} \\over {{{(1 + {e^2})}^2}}}$$"}, {"identifier": "D", "content": "$${{ - {e^{3/2}}} \\over {{{({e^2} + 1)}^2}}}$$"}] | ["D"] | null | $${{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{{{x^2}} \over 2}}} - x} \right)$$<br><br>$$ \Rightarrow {{ - 1} \over {{{(y + 1)}^2}}}{{dy} \over {dx}} - x\left( {{1 \over {y + 1}}} \right) = - {e^{{{{x^2}} \over 2}}}$$<br><br>Put, $${1 \over {y + 1}} = z$$<br><br>$$ - {1 \over {{{(y + 1)}^2}}}.{{dy} \over {dx}} = {{dz} \over {dx}}$$<br><br>$$ \therefore $$ $${{dz} \over {dx}} + z( - x) = - {e^{{{{x^2}} \over 2}}}$$<br><br>$$I.F = {e^{\int { - xdx} }} = {e^{{{ - {x^2}} \over 2}}}$$<br><br>$$z.\left( {{e^{ - {{{x^2}} \over 2}}}} \right) = - \int {{e^{ - {{{x^2}} \over 2}}}.{e^{{{{x^2}} \over 2}}}dx} = - \int {1.dx = - x + C} $$<br><br>$$ \Rightarrow {{{e^{ - {{{x^2}} \over 2}}}} \over {y + 1}} = - x + C$$ .... (1)<br><br>Given y = 0 at x = 2 Put in (1)<br><br>$${{{e^{ - 2}}} \over {0 + 1}} = - 2 + C$$<br><br>$$C = {e^{ - 2}} + 2$$ .... (2)<br><br>From (1) and (2)<br><br>$$y + 1 = {{{e^{ - {{{x^2}} \over 2}}}} \over {{e^{ - 2}} + 2 - x}}$$<br><br>Again, at x = 1<br><br>$$ \Rightarrow y + 1 = {{{e^{3/2}}} \over {{e^2} + 1}}$$<br><br>$$ \Rightarrow y + 1 = {{{e^{3/2}}} \over {{e^2} + 1}}$$<br><br>$$ \therefore $$ $${\left. {{{dy} \over {dx}}} \right|_{x = 1}} = {{{e^{3/2}}} \over {{e^2} + 1}}\left( {{{{e^{3/2}}} \over {{e^2} + 1}} \times {e^{1/2}} - 1} \right)$$<br><br>$$ = - {{{e^{3/2}}} \over {{{({e^2} + 1)}^2}}}$$ | mcq | jee-main-2021-online-18th-march-evening-shift |
1krrtqgzm | maths | differential-equations | linear-differential-equations | Let y = y(x) satisfies the equation $${{dy} \over {dx}} - |A| = 0$$, for all x > 0, where $$A = \left[ {\matrix{
y & {\sin x} & 1 \cr
0 & { - 1} & 1 \cr
2 & 0 & {{1 \over x}} \cr
} } \right]$$. If $$y(\pi ) = \pi + 2$$, then the value of $$y\left( {{\pi \over 2}} \right)$$ is : | [{"identifier": "A", "content": "$${\\pi \\over 2} + {4 \\over \\pi }$$"}, {"identifier": "B", "content": "$${\\pi \\over 2} - {1 \\over \\pi }$$"}, {"identifier": "C", "content": "$${{3\\pi } \\over 2} - {1 \\over \\pi }$$"}, {"identifier": "D", "content": "$${\\pi \\over 2} - {4 \\over \\pi }$$"}] | ["A"] | null | $$|A| = - {y \over x} + 2\sin x + 2$$<br><br>$${{dy} \over {dx}} = |A|$$<br><br>$${{dy} \over {dx}} = - {y \over x} + 2\sin x + 2$$<br><br>$${{dy} \over {dx}} + {y \over x} = 2\sin x + 2$$<br><br>$$I.F. = {e^{\int {{1 \over x}dx} }} = x$$<br><br>$$ \Rightarrow yx = \int {x(2\sin x + 2)dx} $$<br><br>$$xy = {x^2} - 2x\cos x + 2\sin x + c$$ ..... (i)<br><br>Now, x = $$\pi$$, y = $$\pi$$ + 2<br><br>Use in (i)<br><br>c = 0<br><br>Now, (i) becomes<br><br>$$xy = {x^2} - 2x\cos x + 2\sin x$$<br><br>put $$x = \pi /2$$<br><br>$${\pi \over 2}y = {\left( {{\pi \over 2}} \right)^2} - 2.{\pi \over 2}\cos {\pi \over 2} + 2\sin {\pi \over 2}$$<br><br>$$y({\pi \over 2}) = {{{\pi ^2}} \over 4} + 2$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1krtc61w8 | maths | differential-equations | linear-differential-equations | Let y = y(x) be the solution of the differential equation $$\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx$$, with $$y\left( {{\pi \over 4}} \right) = 0$$. Then, the value of $${(y(0) + 1)^2}$$ is equal to : | [{"identifier": "A", "content": "e<sup>1/2</sup>"}, {"identifier": "B", "content": "e<sup>$$-$$1/2</sup>"}, {"identifier": "C", "content": "e<sup>$$-$$1</sup>"}, {"identifier": "D", "content": "e"}] | ["C"] | null | $${{dy} \over {dx}} + 2{\sin ^2}x = 1 + y\cos 2x$$<br><br>$$ \Rightarrow {{dy} \over {dx}} + ( - \cos 2x)y = \cos 2x$$<br><br>$$I.F. = {e^{\int { - \cos 2xdx} }} = {e^{ - {{\sin 2x} \over 2}}}$$<br><br>Solution of D.E.<br><br>$$y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = \int {(\cos 2x)\left( {{e^{ - {{\sin 2x} \over 2}}}} \right)} dx + c$$<br><br>$$ \Rightarrow y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = - {e^{ - {{\sin 2x} \over 2}}} + c$$<br><br>Given<br><br>$$y\left( {{\pi \over 4}} \right) = 0$$<br><br>$$ \Rightarrow 0 = - {e^{{{ - 1} \over 2}}} + c \Rightarrow c = {e^{{{ - 1} \over 2}}}$$<br><br>$$ \Rightarrow y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = - {e^{ - {{\sin 2x} \over 2}}} + {e^{{{ - 1} \over 2}}}$$<br><br>at x = 0<br><br>$$y = - 1 + {e^{{{ - 1} \over 2}}}$$<br><br>$$ \Rightarrow y(0) = - 1 + {e^{{{ - 1} \over 2}}} \Rightarrow {(y(0) + 1)^2} = {e^{ - 1}}$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1krw1vsqv | maths | differential-equations | linear-differential-equations | Let y = y(x) be solution of the following differential equation $${e^y}{{dy} \over {dx}} - 2{e^y}\sin x + \sin x{\cos ^2}x = 0,y\left( {{\pi \over 2}} \right) = 0$$ If $$y(0) = {\log _e}(\alpha + \beta {e^{ - 2}})$$, then $$4(\alpha + \beta )$$ is equal to ______________. | [] | null | 4 | Let e<sup>y</sup> = t<br><br>$$ \Rightarrow {{dt} \over {dx}} - (2\sin x)t = - \sin x{\cos ^2}x$$<br><br>$$I.F. = {e^{2\cos x}}$$<br><br>$$ \Rightarrow t.{e^{2\cos x}} = \int {{e^{2\cos x}}.( - \sin x{{\cos }^2}x)dx} $$<br><br>$$ \Rightarrow {e^y}.{e^{2\cos x}} = \int {{e^{2z}}.{z^2}dz,z = {e^{2\cos x}}} $$<br><br>$$ \Rightarrow {e^y}.{e^{2\cos x}} = {1 \over 2}.{\cos ^2}x.{e^{2\cos x}} - {1 \over 2}\cos x.{e^{2\cos x}} + {{{e^{2\cos x}}} \over 4} + C$$<br><br>at $$x = {\pi \over 2},y = 0 \Rightarrow C = {3 \over 4}$$<br><br>$$ \Rightarrow {e^y} = {1 \over 2}{\cos ^2}x - {1 \over 2}\cos x + {1 \over 4} + {3 \over 4}.{e^{ - 2\cos x}}$$<br><br>$$ \Rightarrow y = \log \left[ {{{{{\cos }^2}x} \over 2} - {{\cos x} \over 2} + {1 \over 4} + {3 \over 4}{e^{ - 2\cos x}}} \right]$$<br><br>Put x = 0<br><br>$$ \Rightarrow y = \log \left[ {{1 \over 4} + {3 \over 4}{e^{ - 2}}} \right] \Rightarrow \alpha = {1 \over 4},\beta = {3 \over 4}$$ | integer | jee-main-2021-online-25th-july-morning-shift |
1ktbbmp8l | maths | differential-equations | linear-differential-equations | Let y = y(x) be a solution curve of the differential equation $$(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$$, $$x \in \left( {0,{\pi \over 2}} \right)$$. If $$\mathop {\lim }\limits_{x \to 0 + } xy(x) = 1$$, then the value of $$y\left( {{\pi \over 4}} \right)$$ is : | [{"identifier": "A", "content": "$$ - {\\pi \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 4} - 1$$"}, {"identifier": "C", "content": "$${\\pi \\over 4} + 1$$"}, {"identifier": "D", "content": "$${\\pi \\over 4}$$"}] | ["D"] | null | $$(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$$<br><br>or $${{dy} \over {dx}} + {{{{\sec }^2}x} \over {\tan x}}.y = - \tan x$$<br><br>$$IF = {e^{\int {{{{{\sec }^2}x} \over {\tan x}}dx} }} = {e^{\ln \tan x}} = \tan x$$<br><br>$$\therefore$$ $$y\tan x = - \int {{{\tan }^2}x\,dx} $$<br><br>or $$y\tan x = - \tan x + x + C$$<br><br>or $$y = - 1 + {x \over {\tan x}} + {C \over {\tan x}}$$<br><br>or $$\mathop {\lim }\limits_{x \to 0} xy = - x + {{{x^2}} \over {\tan x}} + {{Cx} \over {\tan x}} = 1$$<br><br>or C = 1<br><br>$$y(x) = \cot x + x\cot x - 1$$<br><br>$$y\left( {{\pi \over 4}} \right) = {\pi \over 4}$$ | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktelne7i | maths | differential-equations | linear-differential-equations | Let us consider a curve, y = f(x) passing through the point ($$-$$2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x<sup>2</sup>. Then : | [{"identifier": "A", "content": "$${x^2} + 2xf(x) - 12 = 0$$"}, {"identifier": "B", "content": "$${x^3} + xf(x) + 12 = 0$$"}, {"identifier": "C", "content": "$${x^3} - 3xf(x) - 4 = 0$$"}, {"identifier": "D", "content": "$${x^2} + 2xf(x) + 4 = 0$$"}] | ["C"] | null | $$y + {{xdy} \over {dx}} = {x^2}$$ (given)<br><br>$$ \Rightarrow {{dy} \over {dx}} + {y \over x} = x$$<br><br>If $${e^{\int {{1 \over x}dx} }} = x$$<br><br>Solution of DE<br><br>$$ \Rightarrow y\,.\,x = \int {x\,.\,x\,dx} $$<br><br>$$ \Rightarrow xy = {{{x^3}} \over 3} + {c \over 3}$$<br><br>Passes through ($$-$$2, 2), so<br><br>$$-$$12 = $$-$$ 8 + c $$\Rightarrow$$ c = $$-$$ 4<br><br>$$\therefore$$ 3xy = x<sup>3</sup> $$-$$ 4<br><br>i.e. 3x . f(x) = x<sup>3</sup> $$-$$ 4 | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktfzz14c | maths | differential-equations | linear-differential-equations | If the solution curve of the differential equation (2x $$-$$ 10y<sup>3</sup>)dy + ydx = 0, passes through the points (0, 1) and (2, $$\beta$$), then $$\beta$$ is a root of the equation : | [{"identifier": "A", "content": "y<sup>5</sup> $$-$$ 2y $$-$$ 2 = 0"}, {"identifier": "B", "content": "2y<sup>5</sup> $$-$$ 2y $$-$$ 1 = 0"}, {"identifier": "C", "content": "2y<sup>5</sup> $$-$$ y<sup>2</sup> $$-$$ 2 = 0"}, {"identifier": "D", "content": "y<sup>5</sup> $$-$$ y<sup>2</sup> $$-$$ 1 = 0"}] | ["D"] | null | $$(2x - 10{y^3})dy + ydx = 0$$<br><br>$$ \Rightarrow {{dx} \over {dy}} + \left( {{2 \over y}} \right)x = 10{y^2}$$<br><br>$$I.F. = {e^{\int {{2 \over y}dy} }} = {e^{2\ln (y)}} = {y^2}$$<br><br>Solution of D.E. is<br><br>$$\therefore$$ $$x\,.\,y = \int {(10{y^2}){y^2}.\,dy} $$<br><br>$$x{y^2} = {{10{y^5}} \over 5} + C \Rightarrow x{y^2} = 2{y^5} + C$$<br><br>It passes through (0, 1) $$\to$$ 0 = 2 + C $$\Rightarrow$$ C = $$-$$2<br><br>$$\therefore$$ Curve is $$x{y^2} = 2{y^5} - 2$$<br><br>Now, it passes through (2, $$\beta$$)<br><br>$$2{\beta ^2} = 2{\beta ^5} - 2 \Rightarrow {\beta ^5} - {\beta ^2} - 1 = 0$$<br><br>$$\therefore$$ $$\beta$$ is root of an equation $${y^5} - {y^2} - 1 = 0$$ | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktk7ci6b | maths | differential-equations | linear-differential-equations | If $${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$$, y(0) = 0, then for y = 1, the value of x lies in the interval : | [{"identifier": "A", "content": "(1, 2)"}, {"identifier": "B", "content": "$$\\left( {{1 \\over 2},1} \\right]$$"}, {"identifier": "C", "content": "(2, 3)"}, {"identifier": "D", "content": "$$\\left( {0,{1 \\over 2}} \\right]$$"}] | ["A"] | null | $${{dy} \over {dx}} = {{{2^x}(y + {2^y})} \over {{2^x}(1 + {2^y}\ln 2)}}$$<br><br>$$ \Rightarrow \int {{{(1 + {2^y})\ln 2} \over {(y + {2^y})}}dy = \int {dx} } $$<br><br>$$ \Rightarrow \ln \left| {y + {2^y}} \right| = x + c$$<br><br>x = 0; y = 0 $$\Rightarrow$$ c = 0<br><br>$$ \Rightarrow x = \ln \left| {y + {2^y}} \right|$$<br><br>$$\Rightarrow$$ at y = 1, x = ln3<br><br>$$\because$$ $$3 \in (e,{e^2}) \Rightarrow x \in (1,2)$$ | mcq | jee-main-2021-online-31st-august-evening-shift |
1kto2tuh8 | maths | differential-equations | linear-differential-equations | If y = y(x) is the solution curve of the differential equation $${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$$ ; x > 0 and y(1) = 1, then $$y\left( {{1 \over 2}} \right)$$ is equal to : | [{"identifier": "A", "content": "$${3 \\over 2} - {1 \\over {\\sqrt e }}$$"}, {"identifier": "B", "content": "$$3 + {1 \\over {\\sqrt e }}$$"}, {"identifier": "C", "content": "3 + e"}, {"identifier": "D", "content": "3 $$-$$ e"}] | ["D"] | null | $${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$$ ; x > 0, y(1) = 1<br><br>$${x^2}dy + {{(xy - 1)} \over x}dx = 0$$<br><br>$${x^2}dy = {{(xy - 1)} \over x}dx$$<br><br>$${{dy} \over {dx}} = {{1 - xy} \over {{x^3}}}$$<br><br>$${{dy} \over {dx}} = {1 \over {{x^3}}} - {y \over {{x^2}}}$$<br><br>$${{dy} \over {dx}} = {1 \over {{x^2}}}.y = {1 \over {{x^3}}}$$<br><br>If $${e^{\int {{1 \over {{x^2}}}dx} }} = {e^{ - {1 \over x}}}$$<br><br>$$y{e^{ - {1 \over x}}} = \int {{1 \over {{x^3}}}.{e^{ - {1 \over x}}}} $$<br><br>$$y{e^{ - {1 \over x}}} = {e^{ - x}}\left( {1 + {1 \over x}} \right) + C$$<br><br>$$1.\,{e^{ - 1}} = {e^{ - 1}}(2) + C$$<br><br>$$C = - {e^{ - 1}} = - {1 \over e}$$<br><br>$$y{e^{ - {1 \over x}}} = {e^{ - {1 \over x}}}\left( {1 + {1 \over x}} \right) - {1 \over e}$$<br><br>$$y\left( {{1 \over 2}} \right) = 3 - {1 \over e} \times {e^2}$$<br><br>$$y\left( {{1 \over 2}} \right) = 3 - e$$ | mcq | jee-main-2021-online-1st-september-evening-shift |
1l5460552 | maths | differential-equations | linear-differential-equations | <p>Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} + {{\sqrt 2 y} \over {2{{\cos }^4}x - {{\cos }^2}x}} = x{e^{{{\tan }^{ - 1}}(\sqrt 2 \cot 2x)}},\,0 < x < {\pi \over 2}$$ with $$y\left( {{\pi \over 4}} \right) = {{{\pi ^2}} \over {32}}$$. If $$y\left( {{\pi \over 3}} \right) = {{{\pi ^2}} \over {18}}{e^{ - {{\tan }^{ - 1}}(\alpha )}}$$, then the value of 3$$\alpha$$<sup>2</sup> is equal to ___________.</p> | [] | null | 2 | $\frac{d y}{d x}+\frac{2 \sqrt{2} y}{1+\cos ^{2} 2 x}=x e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}$
<br/><br/>
I.F. $=e^{\int \frac{2 \sqrt{2} d x}{1+\cos ^{2} 2 x}}=e^{\sqrt{2} \int \frac{2 \sec ^{2} 2 x}{2+\tan ^{2} 2 x} d x}$ $=e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}$
<br/><br/>
$\Rightarrow y \cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}=\int x e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}$
<br/><br/>
$$
\cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)} d x+c
$$
<br/><br/>
$\Rightarrow y \cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}=e^{\frac{\pi}{2}} \cdot \frac{x^{2}}{2}+c$
<br/><br/>
When $x=\frac{\pi}{4}, y=\frac{\pi^{2}}{32}$ gives $c=0$
<br/><br/>
When $x=\frac{\pi}{3}, y=\frac{\pi^{2}}{18} e^{-\tan ^{-1} \alpha}$
<br/><br/>
So $\frac{\pi^{2}}{18} e^{-\tan ^{-1} \alpha} \cdot e^{-\tan ^{-1}\left(-\sqrt{\left.\frac{3}{2}\right)}\right.}=e^{\pi / 2} \frac{\pi^{2}}{18}$
<br/><br/>
$\Rightarrow \quad-\tan ^{-1} \alpha+\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)=\frac{\pi}{2}$
<br/><br/>
$\Rightarrow \tan ^{-1}(-\alpha)=\tan ^{-1}\left(\sqrt{\frac{2}{3}}\right)$
<br/><br/>
$\Rightarrow \alpha=-\sqrt{\frac{2}{3}} \Rightarrow 3 \alpha^{2}=2$ | integer | jee-main-2022-online-29th-june-morning-shift |
1l54trkcq | maths | differential-equations | linear-differential-equations | <p>Let y = y(x), x > 1, be the solution of the differential equation $$(x - 1){{dy} \over {dx}} + 2xy = {1 \over {x - 1}}$$, with $$y(2) = {{1 + {e^4}} \over {2{e^4}}}$$. If $$y(3) = {{{e^\alpha } + 1} \over {\beta {e^\alpha }}}$$, then the value of $$\alpha + \beta $$ is equal to _________.</p> | [] | null | 14 | $\frac{d y}{d x}+y\left(\frac{2 x}{x-1}\right)=\frac{1}{(x-1)^{2}}$
<br/><br/>
$$
\text { I.F} =e^{\int \frac{2 x}{x-1} d x}
$$
<br/><br/>
$$
\begin{aligned}
& =e^{2 \int\left(\frac{x-1}{x-1}+\frac{1}{x-1}\right) d x} \\\\
& =e^{2 x+2 \ln (x-1)} \\\\
& =\mathrm{e}^{2 x}(\mathrm{x}-1)^{2}
\end{aligned}
$$
<br/><br/>
$\Rightarrow \int d\left(y \cdot e^{2 x}(x-1)^{2}\right)=\int e^{2 x} d x$
<br/><br/>
$\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}}{2}+c$
<br/><br/>
$\downarrow y(2)=\frac{1+e^{4}}{2 e^{4}}$
<br/><br/>
$\frac{1+e^{4}}{2 e^{4}} \cdot e^{4}=\frac{e^{4}}{2}+c$
<br/><br/>
$\Rightarrow \quad c=\frac{e^{4}}{2}\left(\frac{1+e^{4}-e^{4}}{e^{4}}\right)=\frac{1}{2}$
<br/><br/>
$\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}+1}{2}$
<br/><br/>
$$
\downarrow y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}}
$$
<br/><br/>
$\Rightarrow \frac{e^{\alpha}+1}{\beta e^{\alpha}} \cdot e^{6} \cdot 4=\frac{e^{6}+1}{2}$
<br/><br/>
$\Rightarrow \alpha=6$ and $\beta=8 \Rightarrow \alpha+\beta=14$ | integer | jee-main-2022-online-29th-june-evening-shift |
1l55hnalo | maths | differential-equations | linear-differential-equations | <p>Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 $$\tan x(\cos x - y)$$. If the curve passes through the point $$\left( {{\pi \over 4},0} \right)$$, then the value of $$\int\limits_0^{\pi /2} {y\,dx} $$ is equal to :</p> | [{"identifier": "A", "content": "$$(2 - \\sqrt 2 ) + {\\pi \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$$2 - {\\pi \\over {\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$$(2 + \\sqrt 2 ) + {\\pi \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$$2 + {\\pi \\over {\\sqrt 2 }}$$"}] | ["B"] | null | <p>$${{dy} \over {dx}} = 2\tan x(\cos x - y)$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} + 2\tan xy = 2\sin x$$</p>
<p>$$I.F. = {e^{\int {2\tan xdx} }} = {\sec ^2}x$$</p>
<p>$$\therefore$$ Solution of D.E. will be</p>
<p>$$y(x){\sec ^2}x = \int {2\sin x{{\sec }^2}xdx} $$</p>
<p>$$y{\sec ^2}x = 2\sec x + c$$</p>
<p>$$\because$$ Curve passes through $$\left( {{\pi \over 4},0} \right)$$</p>
<p>$$\therefore$$ $$c = - 2\sqrt 2 $$</p>
<p>$$\therefore$$ $$y = 2\cos x - 2\sqrt 2 {\cos ^2}x$$</p>
<p>$$\therefore$$ $$\int_0^{\pi /2} {ydx = \int_0^{\pi /2} {(2\cos x - 2\sqrt 2 {{\cos }^2}x)\,dx} } $$</p>
<p>$$ = 2 - 2\sqrt 2 \,.\,{\pi \over 4} = 2 - {\pi \over {\sqrt 2 }}$$</p> | mcq | jee-main-2022-online-28th-june-evening-shift |
1l566pytt | maths | differential-equations | linear-differential-equations | <p>Let y = y(x) be the solution of the differential equation $$x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0$$, $$x > 1$$, with $$y(2) = - 2$$. Then y(3) is equal to :</p> | [{"identifier": "A", "content": "$$-$$18"}, {"identifier": "B", "content": "$$-$$12"}, {"identifier": "C", "content": "$$-$$6"}, {"identifier": "D", "content": "$$-$$3"}] | ["A"] | null | <p>$${{dy} \over {dx}} + {{y(3{x^2} - 1)} \over {x(1 - {x^2})}} = {{4{x^3}} \over {x(1 - {x^2})}}$$</p>
<p>$$IF = {e^{\int {{{3{x^2} - 1} \over {x - {x^3}}}dx} }} = {e^{ - \ln |{x^3} - x|}} = {e^{ - \ln ({x^3} - x)}} = {1 \over {{x^3} - x}}$$</p>
<p>Solution of D.E. can be given by</p>
<p>$$y.\,{1 \over {{x^3} - x}} = \int {{{4{x^3}} \over {x(1 - {x^2})}}.\,{1 \over {x({x^2} - 1)}}dx} $$</p>
<p>$$ \Rightarrow {y \over {{x^3} - x}} = \int {{{ - 4x} \over {{{({x^2} - 1)}^2}}}dx} $$</p>
<p>$$ \Rightarrow {y \over {{x^3} - x}} = {2 \over {({x^2} - 1)}} + c$$</p>
<p>at x = 2, y = $$-$$2</p>
<p>$${{ - 2} \over 6} = {2 \over 3} + c \Rightarrow c = - 1$$</p>
<p>at $$x = 3 \Rightarrow {y \over {24}} = {2 \over 8} - 1 \Rightarrow y = - 18$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l56r5w5b | maths | differential-equations | linear-differential-equations | <p>If the solution curve of the differential equation <br/><br/>$$(({\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx$$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is</p> | [{"identifier": "A", "content": "2e"}, {"identifier": "B", "content": "$${2 \\over e}$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$${1 \\over e}$$"}] | ["B"] | null | <p>$$\left( {({{\tan }^{ - 1}}y) - x} \right)dy = (1 + {y^2})dx$$</p>
<p>$${{dx} \over {dy}} + {x \over {1 + {y^2}}} = {{{{\tan }^{ - 1}}y} \over {1 + {y^2}}}$$</p>
<p>$$I.F. = {e^{\int {{1 \over {1 + {y^2}}}dy} }} = {e^{{{\tan }^{ - 1}}y}}$$</p>
<p>$$\therefore$$ Solution</p>
<p>$$x.\,{e^{{{\tan }^{ - 1}}y}} = \int {{{{e^{{{\tan }^{ - 1}}y}}{{\tan }^{ - 1}}y} \over {1 + {y^2}}}dy} $$</p>
<p>Let $${e^{{{\tan }^{ - 1}}y}} = t$$</p>
<p>$${{{e^{{{\tan }^{ - 1}}y}}} \over {1 + {y^2}}} = dt$$</p>
<p>$$ = x{e^{{{\tan }^{ - 1}}y}} = \int {\ln tdt = t\ln t - t + c} $$</p>
<p>$$\therefore$$ $$ = x{e^{{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}}{\tan ^{ - 1}}y - {e^{{{\tan }^{ - 1}}y}} + c$$ ..... (i)</p>
<p>$$\because$$ It passes through (1, 0) $$\Rightarrow$$ c = 2</p>
<p>Now put y = tan1, then</p>
<p>$$ex = e - e + 2$$</p>
<p>$$ \Rightarrow x = {2 \over e}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l56u6swe | maths | differential-equations | linear-differential-equations | <p>Let $$y = y(x)$$ be the solution of the differential equation $$(1 - {x^2})dy = \left( {xy + ({x^3} + 2)\sqrt {1 - {x^2}} } \right)dx, - 1 < x < 1$$, and $$y(0) = 0$$. If $$\int_{{{ - 1} \over 2}}^{{1 \over 2}} {\sqrt {1 - {x^2}} y(x)dx = k} $$, then k<sup>$$-$$1</sup> is equal to _____________.</p> | [] | null | 320 | <p>$$\left( {1 - {x^2}} \right)dy = \left( {xy + \left( {{x^3} + 2} \right)\sqrt {1 - {x^2}} } \right)dx$$</p>
<p>$$\therefore$$ $${{dy} \over {dx}} - {x \over {1 - {x^2}}}y = {{{x^3} + 3} \over {\sqrt {1 - {x^2}} }}$$</p>
<p>$$\therefore$$ $$I.F. = {e^{\int { - {x \over {1 - {x^2}}}dx} }} = \sqrt {1 - {x^2}} $$</p>
<p>Solution is</p>
<p>$$y.\,\sqrt {1 - {x^2}} = \int {\left( {{x^3} + 3} \right)dx} $$</p>
<p>$$y.\,\sqrt {1 - {x^2}} = {{{x^4}} \over 4} + 3x + c$$</p>
<p>$$\because$$ $$y(0) = 0 \Rightarrow c = 0$$</p>
<p>$$\therefore$$ $$y(x) = {{{x^4} + 12x} \over {4\sqrt {1 - {x^2}} }}$$</p>
<p>$$\therefore$$ $$\int_{{{ - 1} \over 2}}^{{1 \over 2}} {\sqrt {1 - {x^2}} y(x)dx = \int_{{{ - 1} \over 2}}^{{1 \over 2}} {\left( {{{{x^4} + 12x} \over 4}} \right)dx = \int_0^{{1 \over 2}} {{{{x^4}} \over 2}dx} } } $$</p>
<p>$$\therefore$$ $$k = {1 \over {320}}$$</p>
<p>$$\therefore$$ $$ = {k^{ - 1}} = 320$$</p> | integer | jee-main-2022-online-27th-june-evening-shift |
1l58asnrg | maths | differential-equations | linear-differential-equations | <p>Let $$S = (0,2\pi ) - \left\{ {{\pi \over 2},{{3\pi } \over 4},{{3\pi } \over 2},{{7\pi } \over 4}} \right\}$$. Let $$y = y(x)$$, x $$\in$$ S, be the solution curve of the differential equation $${{dy} \over {dx}} = {1 \over {1 + \sin 2x}},\,y\left( {{\pi \over 4}} \right) = {1 \over 2}$$. If the sum of abscissas of all the points of intersection of the curve y = y(x) with the curve $$y = \sqrt 2 \sin x$$ is $${{k\pi } \over {12}}$$, then k is equal to _____________.</p> | [] | null | 42 | <p>$${{dy} \over {dx}} = {1 \over {1 + \sin 2x}}$$</p>
<p>$$ \Rightarrow dy = {{{{\sec }^2}xdx} \over {{{(1 + \tan x)}^2}}}$$</p>
<p>$$ \Rightarrow y = - {1 \over {1 + \tan x}} + c$$</p>
<p>When $$x = {\pi \over 4}$$, $$y = {1 \over 2}$$ gives c = 1</p>
<p>So $$y = {{\tan x} \over {1 + \tan x}} \Rightarrow y = {{\sin x} \over {\sin x + \cos x}}$$</p>
<p>Now, $$y = \sqrt 2 \sin x \Rightarrow \sin x = 0$$</p>
<p>or $$\sin x + \cos x = {1 \over {\sqrt 2 }}$$</p>
<p>$$\sin x = 0$$ gives $$x = \pi $$ only.</p>
<p>and $$\sin x + \cos x = {1 \over {\sqrt 2 }} \Rightarrow \sin \left( {x + {\pi \over 4}} \right) = {1 \over 2}$$</p>
<p>So $$x + {\pi \over 4} = {{5\pi } \over 6}$$ or $${{13\pi } \over 6} \Rightarrow x = {{7\pi } \over {12}}$$ or $${{23\pi } \over {12}}$$</p>
<p>Sum of all solutions $$ = \pi + {{7\pi } \over {12}} + {{23\pi } \over {12}} = {{42\pi } \over {12}}$$</p>
<p>Hence, $$k = 42$$.</p> | integer | jee-main-2022-online-26th-june-morning-shift |
1l58fd3hx | maths | differential-equations | linear-differential-equations | <p>If $$y = y(x)$$ is the solution of the differential equation <br/><br/>$$x{{dy} \over {dx}} + 2y = x\,{e^x}$$, $$y(1) = 0$$ then the local maximum value <br/><br/>of the function $$z(x) = {x^2}y(x) - {e^x},\,x \in R$$ is :</p> | [{"identifier": "A", "content": "1 $$-$$ e"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${4 \\over e} - e$$"}] | ["D"] | null | <p>$$x{{dy} \over {dx}} + 2y = x{e^x},\,\,y(1) = 0$$</p>
<p>$${{dy} \over {dx}} + {2 \over x}y = {e^x}$$, then $${e^{\int {{2 \over x}dx} }}dx = {x^2}$$</p>
<p>$$y\,.\,{x^2} = \int {{x^2}{e^x}dx} $$</p>
<p>$$y{x^2} = {x^2}{e^x} - \int {2x{e^x}dx} $$</p>
<p>$$ = {x^2}{e^x} - 2(x{e^x} - {e^x}) + c$$</p>
<p>$$y{x^2} = {x^2}{e^x} - 2x{e^x} + 2{e^x} + c$$</p>
<p>$$y{x^2} = ({x^2} - 2x + 2){e^x} + c$$</p>
<p>$$0 = e + c \Rightarrow c = - e$$</p>
<p>$$y(x)\,.\,{x^2} - {e^x} = {(x - 1)^2}{e^x} - e$$</p>
<p>$$z(x) = {(x - 1)^2}{e^x} - e$$</p>
<p>For local maximum $$z'(x) = 0$$</p>
<p>$$\therefore$$ $$2(x - 1){e^x} + {(x - 1)^2}{e^x} = 0$$</p>
<p>$$\therefore$$ $$x = - 1$$</p>
<p>And local maximum value $$ = z( - 1)$$</p>
<p>$$ = {4 \over e} - e$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l58ffh9e | maths | differential-equations | linear-differential-equations | <p>If the solution of the differential equation <br/><br/>$${{dy} \over {dx}} + {e^x}\left( {{x^2} - 2} \right)y = \left( {{x^2} - 2x} \right)\left( {{x^2} - 2} \right){e^{2x}}$$ satisfies $$y(0) = 0$$, then the value of y(2) is _______________.</p> | [{"identifier": "A", "content": "$$-$$1"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "e"}] | ["C"] | null | <p>$$\because$$ $${{dy} \over {dx}} + {e^x}({x^2} - 2)y = ({x^2} - 2x)({x^2} - 2){e^{2x}}$$</p>
<p>Here, $$I.F. = {e^{\int {{e^x}({x^2} - 2)dx} }}$$</p>
<p>$$ = {e^{({x^2} - 2x){e^x}}}$$</p>
<p>$$\therefore$$ Solution of the differential equation is</p>
<p>$$y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {({x^2} - 2x)({x^2} - 2){e^{2x}}\,.\,{e^{({x^2} - 2x){e^x}}}dx} $$</p>
<p>$$ = \int {({x^2} - 2x){e^x}\,.\,({x^2} - 2){e^x}\,.\,{e^{({x^2} - 2x){e^x}}}dx} $$</p>
<p>Let $$({x^2} - 2x){e^x} = t$$</p>
<p>$$\therefore$$ $$({x^2} - 2){e^x}dx = dt$$</p>
<p>$$y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {t\,.\,{e^t}dt} $$</p>
<p>$$y\,.\,{e^{({x^2} - 2x){e^x}}} = ({x^2} - 2x - 1){e^{({x^2} - 2x){e^x}}} + c$$</p>
<p>$$\therefore$$ $$y(0) = 0$$</p>
<p>$$\therefore$$ $$c = 1$$</p>
<p>$$\therefore$$ $$y = ({x^2} - 2x - 1) + {e^{(2x - {x^2}){e^x}}}$$</p>
<p>$$\therefore$$ $$y(2) = - 1 + 1 = 0$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l5aiqgx8 | maths | differential-equations | linear-differential-equations | <p>Let $$y = y(x)$$ be the solution of the differential equation $$(x + 1)y' - y = {e^{3x}}{(x + 1)^2}$$, with $$y(0) = {1 \over 3}$$. Then, the point $$x = - {4 \over 3}$$ for the curve $$y = y(x)$$ is :</p> | [{"identifier": "A", "content": "not a critical point"}, {"identifier": "B", "content": "a point of local minima"}, {"identifier": "C", "content": "a point of local maxima"}, {"identifier": "D", "content": "a point of inflection"}] | ["B"] | null | <p>$$(x + 1){{dy} \over {dx}} - y = {e^{3x}}{(x + 1)^2}$$</p>
<p>$${{dy} \over {dx}} - {y \over {x + 1}} = {e^{3x}}(x + 1)$$</p>
<p>If $${e^{ - \int {{1 \over {x + 1}}x} }} = {e^{ - \log (x + 1)}} = {1 \over {x + 1}}$$</p>
<p>$$\therefore$$ $$y\left( {{1 \over {x + 1}}} \right) = \int {{{{e^{3x}}(x + 1)} \over {x + 1}}dx} $$</p>
<p>$${y \over {x + 1}} = \int {{e^{3x}}dx} $$</p>
<p>$${y \over {x + 1}} = {{{e^{3x}}} \over 3} + c$$</p>
<p>$$\because$$ $$y(0) = {1 \over 3}$$</p>
<p>$${1 \over 3} = {1 \over 3} + c$$</p>
<p>$$\therefore$$ $$c = 0$$</p>
<p>So : $$y = {{{e^{3x}}} \over 3}(x + 1)$$</p>
<p>$$y' = {e^{3x}}(x + 1) + {{{e^{3x}}} \over 3} = {e^{3x}}\left( {x + {4 \over 3}} \right)$$</p>
<p>$$y'' = 3{e^{3x}}\left( {x + {4 \over 3}} \right) + {e^{3x}} = {e^{3x}}(3x + 5)$$</p>
<p>$$y' = 0$$ at $$x = {{ - 4} \over 3}$$ & $$y'' = {e^{ - 4}}(1) > 0$$ at $$x = {{ - 4} \over 3}$$</p>
<p>$$ \Rightarrow x = {{ - 4} \over 3}$$ is point of local minima</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l5c1gfi6 | maths | differential-equations | linear-differential-equations | <p>If x = x(y) is the solution of the differential equation <br/><br/>$$y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0$$; then x(e) is equal to :</p> | [{"identifier": "A", "content": "$${e^3}({e^e} - 1)$$"}, {"identifier": "B", "content": "$${e^e}({e^3} - 1)$$"}, {"identifier": "C", "content": "$${e^2}({e^e} + 1)$$"}, {"identifier": "D", "content": "$${e^e}({e^2} - 1)$$"}] | ["A"] | null | $\frac{d x}{d y}-\frac{2 x}{y}=y^{2}(y+1) e^{y}$
<br/><br/>
$$
\text { If }=e^{\int-\frac{2}{y} d y}=e^{-2 \ln y}=\frac{1}{y^{2}}
$$
<br/><br/>
Solution is given by
<br/><br/>
$$
\begin{aligned}
&x \cdot \frac{1}{y^{2}}=\int y^{2}(y+1) e^{y} \cdot \frac{1}{y^{2}} d y \\\\
\Rightarrow & \frac{x}{y^{2}}=\int(y+1) e^{y} d y \\\\
\Rightarrow & \frac{x}{y^{2}}=y e^{y}+c
\end{aligned}
$$<br/><br/>
$\Rightarrow x=y^{2}\left(y e^{y}+c\right)$ at, $y=1, x=0$
<br/><br/>
$\Rightarrow 0=1\left(1 \cdot e^{1}+c\right) \Rightarrow c=-e$ at $y=e$,
<br/><br/>
$x=e^{2}\left(e . e^{e}-e\right)$ | mcq | jee-main-2022-online-24th-june-morning-shift |
1l6dvizf8 | maths | differential-equations | linear-differential-equations | <p>The slope of the tangent to a curve $$C: y=y(x)$$ at any point $$(x, y)$$ on it is $$\frac{2 \mathrm{e}^{2 x}-6 \mathrm{e}^{-x}+9}{2+9 \mathrm{e}^{-2 x}}$$.
If $$C$$ passes through the points $$\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)$$ and $$\left(\alpha, \frac{1}{2} \mathrm{e}^{2 \alpha}\right)$$, then $$\mathrm{e}^{\alpha}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{3+\\sqrt{2}}{3-\\sqrt{2}}$$"}, {"identifier": "B", "content": "$$\\frac{3}{\\sqrt{2}}\\left(\\frac{3+\\sqrt{2}}{3-\\sqrt{2}}\\right)$$"}, {"identifier": "C", "content": "$$\n\\frac{1}{\\sqrt{2}}\\left(\\frac{\\sqrt{2}+1}{\\sqrt{2}-1}\\right)\n$$"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{2}+1}{\\sqrt{2}-1}$$"}] | ["B"] | null | $\frac{d y}{d x}=\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}=e^{2 x}-\frac{6 e^{-x}}{2+9 e^{-2 x}}$
<br/><br/>
$$
\begin{aligned}
&\int d y=\int e^{2 x} d x-3 \int \underbrace{1+\left(\frac{3 e^{-x}}{\sqrt{2}}\right)^{2}}_{\text {put } e^{-x}=t} d x \\\\
&=\frac{e^{2 x}}{2}+3 \int \frac{d t}{1+\left(\frac{3 t}{\sqrt{2}}\right)^{2}} \\\\
&=\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1} \frac{3 t}{\sqrt{2}}+C
\end{aligned}
$$<br/><br/>
$y=\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3 e^{-x}}{\sqrt{2}}\right)+C$
<br/><br/>
It is given that the curve passes through
<br/><br/>
$$
\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)
$$
<br/><br/>
$$
\begin{aligned}
& \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}=\frac{1}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)+C
\end{aligned}
$$
<br/><br/>
$\Rightarrow \quad C=\frac{\pi}{2 \sqrt{2}}-\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)$
<br/><br/>
Now if $\left(\alpha, \frac{1}{2} e^{2 \alpha}\right)$ satisfies the curve, then
<br/><br/>
$$
\frac{1}{2} e^{2 \alpha}=\frac{e^{2 \alpha}}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3 e^{-\alpha}}{\sqrt{2}}\right)+\frac{\pi}{2 \sqrt{2}}-\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)
$$
<br/><br/>
$\tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)-\tan ^{-1}\left(\frac{3 e^{-\alpha}}{\sqrt{2}}\right)=\frac{\pi}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{\pi}{4}$
<br/><br/>
$$
\frac{\frac{3}{\sqrt{2}}-\frac{3 e^{-\alpha}}{\sqrt{2}}}{1+\frac{9}{2} e^{-\alpha}}=1
$$
<br/><br/>
$$
\frac{3}{\sqrt{2}} e^{\alpha}-\frac{3}{\sqrt{2}}=e^{\alpha}+\frac{9}{2}
$$
<br/><br/>
$$
e^{\alpha}=\frac{\frac{9}{2}+\frac{3}{\sqrt{2}}}{\frac{3}{\sqrt{2}}-1}=\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)
$$ | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6gifynu | maths | differential-equations | linear-differential-equations | <p>If $${{dy} \over {dx}} + 2y\tan x = \sin x,\,0 < x < {\pi \over 2}$$ and $$y\left( {{\pi \over 3}} \right) = 0$$, then the maximum value of $$y(x)$$ is :</p> | [{"identifier": "A", "content": "$${1 \\over 8}$$"}, {"identifier": "B", "content": "$${3 \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$${3 \\over 8}$$"}] | ["A"] | null | <p>$${{dy} \over {dx}} + 2y\tan x = \sin x$$</p>
<p>which is a first order linear differential equation.</p>
<p>Integrating factor (I. F.) $$ = {e^{\int {2\tan x\,dx} }}$$</p>
<p>$$ = {e^{2\ln |\sec x|}} = {\sec ^2}x$$</p>
<p>Solution of differential equation can be written as</p>
<p>$$y\,.\,{\sec ^2}x = \int {\sin x\,.\,{{\sec }^2}x\,dx = \int {\sec \,x\,.\,\tan x\,dx} } $$</p>
<p>$$y~{\sec ^2}x = \sec x + C$$</p>
<p>$$y\left( {{\pi \over 3}} \right) = 0,0 = \sec {\pi \over 3} + C \Rightarrow \,\,\,\,C = - 2$$</p>
<p>$$y = {{\sec x - 2} \over {{{\sec }^2}x}} = \cos x - 2{\cos ^2}x$$</p>
<p>$$ = {1 \over 8} - 2{\left( {\cos x - {1 \over 4}} \right)^2}$$</p>
<p>$${y_{\max }} = {1 \over 8}$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6hyo0c6 | maths | differential-equations | linear-differential-equations | <p>Let the solution curve $$y=f(x)$$ of the differential equation $$
\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}$$, $$x\in(-1,1)$$ pass through the origin. Then $$\int\limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x
$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{3}-\\frac{1}{4}$$"}, {"identifier": "B", "content": "$$\\frac{\\pi}{3}-\\frac{\\sqrt{3}}{4}$$"}, {"identifier": "C", "content": "$$\\frac{\\pi}{6}-\\frac{\\sqrt{3}}{4}$$"}, {"identifier": "D", "content": "$$\\frac{\\pi}{6}-\\frac{\\sqrt{3}}{2}$$"}] | ["B"] | null | <p>$${{dy} \over {dx}} + {{xy} \over {{x^2} - 1}} = {{{x^4} + 2x} \over {\sqrt {1 - {x^2}} }}$$</p>
<p>which is first order linear differential equation.</p>
<p>Integrating factor $$(I.F.) = {e^{\int {{x \over {{x^2} - 1}}dx} }}$$</p>
<p>$$ = {e^{{1 \over 2}\ln |{x^2} - 1|}} = \sqrt {|{x^2} - 1|} $$</p>
<p>$$ = \sqrt {1 - {x^2}} $$</p>
<p>$$\because$$ $$x \in ( - 1,1)$$</p>
<p>Solution of differential equation</p>
<p>$$y\sqrt {1 - {x^2}} = \int {({x^4} + 2x)dx = {{{x^5}} \over 5} + {x^2} + c} $$</p>
<p>Curve is passing through origin, $$c = 0$$</p>
<p>$$y = {{{x^5} + 5{x^2}} \over {5\sqrt {1 - {x^2}} }}$$</p>
<p>$$\int\limits_{{{ - \sqrt 3 } \over 2}}^{{{\sqrt 3 } \over 2}} {{{{x^5} + 5{x^2}} \over {5\sqrt {1 - {x^2}} }}dx = 0 + 2\int\limits_0^{{{\sqrt 3 } \over 2}} {{{{x^2}} \over {\sqrt {1 - {x^2}} }}dx} } $$</p>
<p>put $$x = \sin \theta $$</p>
<p>$$dx = \cos \theta \,d\theta $$</p>
<p>$$I = 2\int\limits_0^{{\pi \over 3}} {{{{{\sin }^2}\theta \,.\,\cos \theta d\theta } \over {\cos \theta }}} $$</p>
<p>$$ = \int\limits_0^{{\pi \over 3}} {(1 - \cos 2\theta )d\theta } $$</p>
<p>$$ = \left. {\left( {\theta - {{\sin 2\theta } \over 2}} \right)} \right|_0^{{\pi \over 3}}$$</p>
<p>$$ = {\pi \over 3} - {{\sqrt 3 } \over 4}$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6i02qdj | maths | differential-equations | linear-differential-equations | <p>Suppose $$y=y(x)$$ be the solution curve to the differential equation $$\frac{d y}{d x}-y=2-e^{-x}$$ such that $$\lim\limits_{x \rightarrow \infty} y(x)$$ is finite. If $$a$$ and $$b$$ are respectively the $$x$$ - and $$y$$-intercepts of the tangent to the curve at $$x=0$$, then the value of $$a-4 b$$ is equal to _____________.</p> | [] | null | 3 | <p>IF $$ = {e^{-x}}$$</p>
<p>$$y\,.\,{e^{-x}} = - 2{e^{ - x}} + {{{e^{ - 2x}}} \over 2} + C$$</p>
<p>$$ \Rightarrow y = - 2 + {e^{ - x}} + C{e^x}$$</p>
<p>$$\mathop {\lim }\limits_{x \to \infty } \,y(x)$$ is finite so $$C = 0$$</p>
<p>$$y = - 2 + {e^{ - x}}$$</p>
<p>$$ \Rightarrow {\left. {{{dy} \over {dx}} = - {e^{ - x}} \Rightarrow {{dy} \over {dx}}} \right|_{x = 0}} = - 1$$</p>
<p>Equation of tangent</p>
<p>$$y + 1 = - 1(x - 0)$$</p>
<p>or $$y + x = - 1$$</p>
<p>So $$a = - 1,\,b = - 1$$</p>
<p>$$ \Rightarrow a - 4b = 3$$</p> | integer | jee-main-2022-online-26th-july-evening-shift |
1l6jbvr6c | maths | differential-equations | linear-differential-equations | <p>Let $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ be two distinct solutions of the differential equation $$\frac{d y}{d x}=x+y$$, with $$y_{1}(0)=0$$ and $$y_{2}(0)=1$$ respectively. Then, the number of points of intersection of $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ is</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["A"] | null | <p>$${{dy} \over {dx}} = x + y$$</p>
<p>Let $$x + y = t$$</p>
<p>$$1 + {{dy} \over {dx}} = {{dt} \over {dx}}$$</p>
<p>$${{dt} \over {dx}} - 1 = t \Rightarrow \int {{{dt} \over {t + 1}} = \int {dx} } $$</p>
<p>$$\ln |t + 1| = x + C'$$</p>
<p>$$|t + 1| = C{e^x}$$</p>
<p>$$|x + y + 1| = C{e^x}$$</p>
<p>For $${y_1}(x),\,{y_1}(0) = 0 \Rightarrow C = 1$$</p>
<p>For $${y_2}(x),\,{y_2}(0) = 1 \Rightarrow C = 2$$</p>
<p>$${y_1}(x)$$ is given by $$|x + y + 1| = {e^x}$$</p>
<p>$${y_2}(x)$$ is given by $$|x + y + 1| = 2{e^x}$$</p>
<p>At point of intersection</p>
<p>$${e^x} = 2{e^x}$$</p>
<p>No solution</p>
<p>So, there is no point of intersection of $${y_1}(x)$$ and $${y_2}(x)$$.</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6jeco5d | maths | differential-equations | linear-differential-equations | <p>Let $$y=y(x)$$ be the solution curve of the differential equation</p>
<p>$$\sin \left( {2{x^2}} \right){\log _e}\left( {\tan {x^2}} \right)dy + \left( {4xy - 4\sqrt 2 x\sin \left( {{x^2} - {\pi \over 4}} \right)} \right)dx = 0$$, $$0 < x < \sqrt {{\pi \over 2}} $$, which passes through the point $$\left(\sqrt{\frac{\pi}{6}}, 1\right)$$. Then $$\left|y\left(\sqrt{\frac{\pi}{3}}\right)\right|$$ is equal to ______________.</p> | [] | null | 1 | <p>$${{dy} \over {dx}} + y\left( {{{4x} \over {\sin (2{x^2})\ln (\tan {x^2})}}} \right) = {{4\sqrt 2 x\sin \left( {{x^2} - {\pi \over 4}} \right)} \over {\sin (2{x^2})\ln (\tan {x^2})}}$$</p>
<p>$$I.F = {e^{\int {{{4x} \over {\sin (2{x^2})\ln (\tan {x^2})}}dx} }}$$</p>
<p>$$ = {e^{\ln |\ln (\tan {x^2})}} = \ln (\tan {x^2})$$</p>
<p>$$\therefore$$ $$\int {d(y.\ln (\tan {x^2})) = \int {{{4\sqrt 2 x\sin \left( {{x^2} - {\pi \over 4}} \right)} \over {\sin (2{x^2})}}dx} } $$</p>
<p>$$ \Rightarrow y\ln (\tan {x^2}) = \ln \left| {{{\sec {x^2} + \tan {x^2}} \over {{\mathop{\rm cosec}\nolimits} \,{x^2} - \cot {x^2}}}} \right| + C$$</p>
<p>$$\ln \left( {{1 \over {\sqrt 3 }}} \right) = \ln \left( {{{{3 \over {\sqrt 3 }}} \over {2 - \sqrt 3 }}} \right) + C$$</p>
<p>$$e = \ln \left( {{1 \over {\sqrt 3 }}} \right) - \ln \left( {{{\sqrt 3 } \over {2 - \sqrt 3 }}} \right)$$</p>
<p>For $$y\left( {\sqrt {{\pi \over 3}} } \right)$$</p>
<p>$$y\ln \left( {\sqrt 3 } \right) = \ln \left| {{{2 + \sqrt 3 } \over {{2 \over {\sqrt 3 }} + {1 \over {\sqrt 3 }}}}} \right| + \ln \left( {{1 \over {\sqrt 3 }}} \right) - \ln \left( {{{\sqrt 3 } \over {2\sqrt 3 }}} \right)$$</p>
<p>$$ = \ln \left( {2 + \sqrt 3 } \right) + \ln \left( {{1 \over {\sqrt 3 }}} \right) + \ln \left( {{1 \over {\sqrt 3 }}} \right) - \ln \left( {{{\sqrt 3 } \over {2 - \sqrt 3 }}} \right)$$</p>
<p>$$ \Rightarrow y\ln \sqrt 3 = \ln \left( {{1 \over {\sqrt 3 }}} \right)$$</p>
<p>$$ \Rightarrow {y \over 2}\ln 3 = - {1 \over 2}\ln 3$$</p>
<p>$$ \Rightarrow y = 1$$</p>
<p>$$\therefore$$ $$\left| {y\left( {\sqrt {{\pi \over 3}} } \right)} \right| = 1$$.</p> | integer | jee-main-2022-online-27th-july-morning-shift |
1l6nmgd64 | maths | differential-equations | linear-differential-equations | <p>Let $$y=y(x)$$ be the solution curve of the differential equation $$
\frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}$$, $$x >1$$ passing through the point $$\left(2, \sqrt{\frac{1}{3}}\right)$$. Then $$\sqrt{7}\, y(8)$$ is equal to :</p> | [{"identifier": "A", "content": "$$11+6 \\log _{e} 3$$"}, {"identifier": "B", "content": "19"}, {"identifier": "C", "content": "$$12-2 \\log _{\\mathrm{e}} 3$$"}, {"identifier": "D", "content": "$$19-6 \\log _{\\mathrm{e}} 3$$"}] | ["D"] | null | <p>$${{dy} \over {dx}} + {1 \over {{x^2} - 1}}y = \sqrt {{{x - 1} \over {x + 1}}} ,\,x > 1$$</p>
<p>Integrating factor I.F. $$ = {e^{\int {{1 \over {{x^2} - 1}}dx} }} = {e^{{1 \over 2}\ln \left| {{{x - 1} \over {x + 1}}} \right|}}$$</p>
<p>$$ = \sqrt {{{x - 1} \over {x + 1}}} $$</p>
<p>Solution of differential equation</p>
<p>$$y\sqrt {{{x - 1} \over {x + 1}}} = \int {{{x - 1} \over {x + 1}}dx = \int {\left( {1 - {2 \over {x + 1}}} \right)dx} } $$</p>
<p>$$y\sqrt {{{x - 1} \over {x + 1}}} = x - 2\ln |x + 1| + C$$</p>
<p>Curve passes through $$\left( {2,\sqrt {{1 \over 3}} } \right)$$</p>
<p>$${1 \over {\sqrt 3 }} \times {1 \over {\sqrt 3 }} = 2 - 2\ln 3 + C$$</p>
<p>$$C = 2\ln 3 - {5 \over 3}$$</p>
<p>$$y(8) \times {{\sqrt 7 } \over 3} = 8 - 2\ln 9 + 2\ln 3 - {5 \over 3}$$</p>
<p>$$\sqrt 7 \,.\,y(8) = 19 - 6\ln 3$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1l6p1qk2v | maths | differential-equations | linear-differential-equations | <p>Let the solution curve $$y=y(x)$$ of the differential equation $$\left(1+\mathrm{e}^{2 x}\right)\left(\frac{\mathrm{d} y}{\mathrm{~d} x}+y\right)=1$$ pass through the point $$\left(0, \frac{\pi}{2}\right)$$. Then, $$\lim\limits_{x \rightarrow \infty} \mathrm{e}^{x} y(x)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\n\\frac{\\pi}{4}\n$$"}, {"identifier": "B", "content": "$$\n\\frac{3\\pi}{4}\n$$"}, {"identifier": "C", "content": "$$\n\\frac{\\pi}{2}\n$$"}, {"identifier": "D", "content": "$$\n\\frac{3\\pi}{2}\n$$"}] | ["B"] | null | <p>D.E. $$(1 + {e^{2x}})\left( {{{dy} \over {dx}} + y} \right) = 1$$</p>
<p>$$ \Rightarrow {{dy} \over {dx}} + y = {1 \over {1 + {e^{2x}}}}$$</p>
<p>I.F. $$ = {e^{\int {1\,.\,dx} }} = {e^x}$$</p>
<p>$$\therefore$$ Solution</p>
<p>$${e^x}y(x) = \int {{{{e^x}} \over {1 + {e^{2x}}}}dx} $$</p>
<p>$$ \Rightarrow {e^x}y(x) = {\tan ^{ - 1}}({e^x}) + C$$</p>
<p>$$\because$$ It passes through $$\left( {0,{\pi \over 2}} \right),\,C = {\pi \over 2} - {\pi \over 4} = {\pi \over 4}$$</p>
<p>$$\therefore$$ $$\mathop {\lim }\limits_{x \to \infty } {e^x}y(x) = \mathop {\lim }\limits_{x \to \infty } {\tan ^{ - 1}}({e^x}) + {\pi \over 4}$$</p>
<p>$$ = {{3\pi } \over 4}$$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
1ldo6zp5b | maths | differential-equations | linear-differential-equations | <p>Let $$\alpha x=\exp \left(x^{\beta} y^{\gamma}\right)$$ be the solution of the differential equation $$2 x^{2} y \mathrm{~d} y-\left(1-x y^{2}\right) \mathrm{d} x=0, x > 0,y(2)=\sqrt{\log _{e} 2}$$. Then $$\alpha+\beta-\gamma$$ equals :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$-1$$"}] | ["A"] | null | $\begin{aligned} & 2 x^2 y d y-\left(1-x y^2\right) d x=0 \\\\ & \Rightarrow 2 x^2 y \frac{d y}{d x}-1+x y^2=0 \\\\ & \Rightarrow 2 y \frac{d y}{d x}-\frac{1}{x^2}+\frac{y^2}{x}=0 \\\\ & \Rightarrow 2 y \frac{d y}{d x}+\frac{y^2}{x}=\frac{1}{x^2} \text { (L.D.E) } \\\\ & Let, y^2=t \Rightarrow 2 y \frac{d y}{d x}=\frac{d t}{d x} \\\\ & \frac{d t}{d x}+\frac{t}{x}=\frac{1}{x^2} \\\\ & \text { I.F }=e^{\int \frac{1}{x} d x}=x\end{aligned}$
<br/><br/>So, the general solution is
<br/><br/>$$
\begin{aligned}
& \Rightarrow t \times x=\int x \cdot \frac{1}{x^2} d x \\\\
& \Rightarrow y^2 \cdot x=\ln x+c \\\\
& \text { Also,y(2) }=\sqrt{\log _e 2} \\\\
& \log _e^2 2=\log _e 2+c \Rightarrow c=\log _e 2 \\\\
& \Rightarrow y^2 x=\ln x+\ln 2 \\\\
& \Rightarrow y^2 x=\ln 2 x \\\\
& 2 x=\exp \left(x^1 y^2\right)
\end{aligned}
$$
<br/><br/>By compare it with given solution we get,
<br/><br/>$$
\begin{aligned}
& \alpha=2,\beta=1, \gamma=2 \\\\
& \Rightarrow \alpha+\beta-\gamma=2+1-2=1
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-evening-shift |
1ldonlzbl | maths | differential-equations | linear-differential-equations | <p>If $$y=y(x)$$ is the solution curve of the differential equation <br/><br/>$$\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$$, then $$y\left(\frac{\pi}{6}\right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{12}-\\frac{\\sqrt{3}}{2} \\log _{e}\\left(\\frac{2 \\sqrt{3}}{e}\\right)$$"}, {"identifier": "B", "content": "$$\\frac{\\pi}{12}+\\frac{\\sqrt{3}}{2} \\log _{e}\\left(\\frac{2 \\sqrt{3}}{e}\\right)$$"}, {"identifier": "C", "content": "$$\\frac{\\pi}{12}+\\frac{\\sqrt{3}}{2} \\log _{e}\\left(\\frac{2}{e \\sqrt{3}}\\right)$$"}, {"identifier": "D", "content": "$$\\frac{\\pi}{12}-\\frac{\\sqrt{3}}{2} \\log _{e}\\left(\\frac{2}{e \\sqrt{3}}\\right)$$"}] | ["D"] | null | $\frac{d y}{d x}+y \tan x=x \sec x$
<br/><br/>$\therefore $ I.F $=e^{\int \tan x d x}=\sec x$
<br/><br/>$\Rightarrow y \sec x=\int x \sec ^{2} x d x$
<br/><br/>$\Rightarrow y \sec x=x \tan x-\ln |\sec x|+c$
<br/><br/>Given, $$
y(0)=1
$$
<br/><br/>$\Rightarrow 1=c$
<br/><br/>$$ \therefore $$ $ y \sec x=x \tan x-\ln |\sec x|+1$
<br/><br/>$\Rightarrow y=x \sin x-\cos x \ln |\sec x|+cosx$
<br/><br/>Now, at $x=\pi / 6$, we have
<br/><br/>$$
\begin{aligned}
& y=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e \frac{2}{\sqrt{3}}+\frac{\sqrt{3}}{2} \\\\
& =\frac{\pi}{12}-\frac{\sqrt{3}}{2}\left[\log _e \frac{2}{\sqrt{3}}-\log _e e\right]\\\\
&=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right)
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldr6om0l | maths | differential-equations | linear-differential-equations | <p>Let the solution curve $$y=y(x)$$ of the differential equation <br/><br/>$$
\frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}} y=2 x \exp \left\{\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{\left(1+x^{6}\right)}}\right\} \text { pass through the origin. Then } y(1) \text { is equal to : }
$$</p> | [{"identifier": "A", "content": "$$\\exp \\left(\\frac{1-\\pi}{4 \\sqrt{2}}\\right)$$"}, {"identifier": "B", "content": "$$\\exp \\left(\\frac{4-\\pi}{4 \\sqrt{2}}\\right)$$"}, {"identifier": "C", "content": "$$\\exp \\left(\\frac{4+\\pi}{4 \\sqrt{2}}\\right)$$"}, {"identifier": "D", "content": "$$\\exp \\left(\\frac{\\pi-4}{4 \\sqrt{2}}\\right)$$"}] | ["B"] | null | <p>$${{dy} \over {dx}} - {{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}y = 2x\exp \left\{ {{{{x^3} - {{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }}} \right\}$$</p>
<p>$$IF = {e^{ - \int {{{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}dx} }}$$</p>
<p>Let $${\tan ^{ - 1}}{x^3} = t \Rightarrow {{3{x^2}} \over {1 + {x^6}}}dx = dt$$</p>
<p>$$ \Rightarrow IF = {e^{ - \int {{{\tan t} \over {\sec t}}.t\,dt} }} = {e^{ - \int {\sin t.tdt} }} = {e^{t\cos t - \sin t}}$$</p>
<p>$$ \Rightarrow IF = {e^{{{{{\tan }^{ - 1}}({x^3})} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}}$$</p>
<p>$$\therefore$$ Solution is</p>
<p>$$y\,.\,{e^{{{{{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}} = \int {2x\,dx + c} $$</p>
<p>$$ \Rightarrow y\,.\,{e^{{{{{\tan }^{ - 1}}{x^3} - {x^3}} \over {\sqrt {1 + {x^6}} }}}} = {x^2} + c$$</p>
<p>$$y(0) = 0 \Rightarrow c = 0$$</p>
<p>$$x = 1$$</p>
<p>$$y\,.\,{e^{{{{\pi \over 4} - 1} \over {\sqrt 2 }}}} = 1$$</p>
<p>$$ \Rightarrow y = {e^{{{1 - {\pi \over 4}} \over {\sqrt 2 }}}}$$</p>
<p>$$ \Rightarrow y = {e^{{{4 - \pi } \over {4\sqrt 2 }}}}$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldsej5cp | maths | differential-equations | linear-differential-equations | <p>Let $$y=y(x)$$ be the solution of the differential equation $$x{\log _e}x{{dy} \over {dx}} + y = {x^2}{\log _e}x,(x > 1)$$. If $$y(2) = 2$$, then $$y(e)$$ is equal to</p> | [{"identifier": "A", "content": "$${{1 + {e^2}} \\over 2}$$"}, {"identifier": "B", "content": "$${{1 + {e^2}} \\over 4}$$"}, {"identifier": "C", "content": "$${{2 + {e^2}} \\over 2}$$"}, {"identifier": "D", "content": "$${{4 + {e^2}} \\over 4}$$"}] | ["D"] | null | <p>$$x\ln x{{dy} \over {dx}} + y = {x^2}\ln x$$</p>
<p>$${{dy} \over {dx}} + {1 \over {x\ln x}}\,.\,y = x$$</p>
<p>If $$ = {e^{\int {{1 \over {x\ln x}}dx} }} = {e^{\int {{1 \over t}dt} }}$$, where $$t = \ln x$$</p>
<p>$$ = {e^{\ln t}} = t = \ln x$$</p>
<p>$$y.\ln x = \int {x\ln x = {{{x^2}} \over 2}\ln x - \int {{{{x^2}} \over 2}\,.\,{1 \over x}} } $$</p>
<p>$$y\ln x = {{{x^2}} \over 2}\ln x - {{{x^2}} \over 4} + C$$ ..... (i)</p>
<p>$$y(2) = 2 \Rightarrow C = 1$$</p>
<p>Putting $$x = e$$ in (i),</p>
<p>$$y = {{{e^2}} \over 4} + 1 = {{4 + {e^2}} \over 4}$$</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldu5k41z | maths | differential-equations | linear-differential-equations | <p>Let $$y=y(t)$$ be a solution of the differential equation $${{dy} \over {dt}} + \alpha y = \gamma {e^{ - \beta t}}$$ where, $$\alpha > 0,\beta > 0$$ and $$\gamma > 0$$. Then $$\mathop {\lim }\limits_{t \to \infty } y(t)$$</p> | [{"identifier": "A", "content": "is 0"}, {"identifier": "B", "content": "is 1"}, {"identifier": "C", "content": "is $$-1$$"}, {"identifier": "D", "content": "does not exist"}] | ["A"] | null | $\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}$
<br/><br/>
$$
\begin{aligned}
& \text { I.F. }=e^{\int \alpha d t}=e^{\alpha t} \\\\
& \Rightarrow y \cdot e^{\alpha t}=\gamma \int e^{(\alpha-\beta) t} d t=\gamma \frac{e^{(\alpha-\beta) t}}{(\alpha-\beta)}+C \\\\
& \Rightarrow y=\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t} \\\\
& \lim _{x \rightarrow \infty} y(t)=\lim _{x \rightarrow \infty}\left[\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t}\right]=0
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldv2gz1b | maths | differential-equations | linear-differential-equations | <p>Let $$y = y(x)$$ be the solution curve of the differential equation $${{dy} \over {dx}} = {y \over x}\left( {1 + x{y^2}(1 + {{\log }_e}x)} \right),x > 0,y(1) = 3$$. Then $${{{y^2}(x)} \over 9}$$ is equal to :</p> | [{"identifier": "A", "content": "$${{{x^2}} \\over {5 - 2{x^3}(2 + {{\\log }_e}{x^3})}}$$"}, {"identifier": "B", "content": "$${{{x^2}} \\over {3{x^3}(1 + {{\\log }_e}{x^2}) - 2}}$$"}, {"identifier": "C", "content": "$${{{x^2}} \\over {7 - 3{x^3}(2 + {{\\log }_e}{x^2})}}$$"}, {"identifier": "D", "content": "$${{{x^2}} \\over {2{x^3}(2 + {{\\log }_e}{x^3}) - 3}}$$"}] | ["A"] | null | $$
\begin{aligned}
& \frac{d y}{d x}-\frac{y}{x}=y^3\left(1+\log _e x\right) \\\\
& \frac{1}{y^3} \frac{d y}{d x}-\frac{1}{x y^2}=1+\log _e x \\\\
& \text { Let }-\frac{1}{y^2}=t \Rightarrow \frac{2}{y^3} \frac{d y}{d x}=\frac{d t}{d x} \\\\
& \therefore \frac{d t}{d x}+\frac{2 t}{x}=2\left(1+\log _e x\right) \\\\
& \text { I.F. }=e^{\int \frac{2}{x} d x}=x^2 \\\\
& \frac{-x^2}{y^2}=\frac{2}{3}\left(\left(1+\log _e x\right) x^3-\frac{x^3}{3}\right)+C \\\\
& y(1)=3 \\\\
& \frac{y^2}{9}=\frac{C}{5-2 x^3\left(2+\log _e x^3\right)}
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldyb0ayc | maths | differential-equations | linear-differential-equations | <p>Let $$y = y(x)$$ be the solution of the differential equation $${x^3}dy + (xy - 1)dx = 0,x > 0,y\left( {{1 \over 2}} \right) = 3 - \mathrm{e}$$. Then y (1) is equal to</p> | [{"identifier": "A", "content": "2 $$-$$ e"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "e"}] | ["C"] | null | $x^{3} d y+x y d x-d x=0$
<br/><br/>
$\Rightarrow \frac{d y}{d x}=\frac{1-x y}{x^{3}}$
<br/><br/>
$\Rightarrow \frac{d y}{d x}+\frac{y}{x^{2}}=\frac{1}{x^{3}}$
<br/><br/>
I.F. $=e^{\int \frac{d x}{x^{2}}}=e^{-\frac{1}{x}}$
<br/><br/>
$\therefore \quad y e^{-\frac{1}{x}}=\int \frac{e^{-\frac{1}{x}}}{x^{3}} d x$
<br/><br/>
For RHS put $-\frac{1}{x}=t \Rightarrow \frac{d x}{x^{2}}=d t$
<br/><br/>
$\therefore y e^{-\frac{1}{x}}=-\int t e^{t} d t$
<br/><br/>
$\Rightarrow y e^{-\frac{1}{x}}=-\left[t e^{t}-e^{t}\right]+c$
<br/><br/>
$\Rightarrow y e^{-\frac{1}{x}}=\frac{e^{-\frac{1}{x}}}{x}+e^{-\frac{1}{x}}+c$
<br/><br/>
$$
\downarrow y\left(\frac{1}{2}\right)=3-e
$$
<br/><br/>
$\Rightarrow(3-e) e^{-2}=2 e^{-2}+e^{-2}+c$
<br/><br/>
$\Rightarrow \quad c=-\frac{1}{e}$
<br/><br/>
For $y(1)$ put $x=1, c=-e^{-1}$ in equation (i) we get $y e^{-1}=e^{-1}+e^{-1}-e^{-1}$
<br/><br/>
$\Rightarrow y=1$ | mcq | jee-main-2023-online-24th-january-morning-shift |
lgnwbvoh | maths | differential-equations | linear-differential-equations | Let $x=x(y)$ be the solution of the differential equation
<br/><br/>$2(y+2) \log _{e}(y+2) d x+\left(x+4-2 \log _{e}(y+2)\right) d y=0, y>-1$
<br/><br/>with $x\left(e^{4}-2\right)=1$. Then $x\left(e^{9}-2\right)$ is equal to : | [{"identifier": "A", "content": "$\\frac{4}{9}$"}, {"identifier": "B", "content": "$\\frac{32}{9}$"}, {"identifier": "C", "content": "$\\frac{10}{3}$\n"}, {"identifier": "D", "content": "3"}] | ["B"] | null | $$
\begin{aligned}
& 2(y+2) \ln (y+2) d x+(x+4-2 \ln (y+2)) d y=0 \\\\
& 2 \ln (y+2)+(x+4-2 \ln (y+2)) \frac{1}{y+2} \cdot \frac{d y}{d x}=0 \\\\
& \text { let, } \ln (y+2)=t \\\\
& \frac{1}{y+2} \cdot \frac{d y}{d x}=\frac{d t}{d x} \\\\
& 2 t+(x+4-2 t) \cdot \frac{d t}{d x}=0 \\\\
& (x+4-2 t) \frac{d t}{d x}=-2 t \\\\
& \frac{d x}{d t}=\frac{2 t-4-x}{2 t} \\\\
& \frac{d x}{d t}+\frac{x}{2 t}=\frac{2 t-4}{2 t}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& x \cdot t^{1 / 2}=\int \frac{2 t-4}{2 t} \cdot t^{1 / 2} \cdot d t \\\\
& x \cdot t^{1 / 2}=\int\left(t^{1 / 2}-\frac{2}{t^{1 / 2}}\right) \cdot d t \\\\
& =\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-2 \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C \\\\
& x \cdot t^{\frac{1}{2}}=\frac{2 t^{\frac{3}{2}}}{3}-4 \mathrm{t}^{\frac{1}{2}}+C \\\\
& x=\frac{2}{3} \cdot \mathrm{t}-4+\mathrm{C} \cdot \mathrm{t}^{\frac{-1}{2}} \\\\
& x=\frac{2}{3} \ln (\mathrm{y}+2)-4+C \cdot(\ln (\mathrm{y}+2))^{\frac{-1}{2}}
\end{aligned}
$$
<br/><br/>Put $y=e^4-2, x=1$
<br/><br/>$$
\begin{aligned}
& 1=\frac{2}{3} \times 4-4+C \times \frac{1}{2} \\\\
& \frac{C}{2}=5-\frac{8}{3}=\frac{7}{3} \\\\
& C=\frac{14}{3} \\\\
& x=\frac{2}{3} \times 9-4+\frac{14}{3} \times \frac{1}{3} \\\\
& =2+\frac{14}{9} \\\\
& =\frac{32}{9}
\end{aligned}
$$ | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgoy5uje | maths | differential-equations | linear-differential-equations | <p>If $$y=y(x)$$ is the solution of the differential equation <br/><br/>$$\frac{d y}{d x}+\frac{4 x}{\left(x^{2}-1\right)} y=\frac{x+2}{\left(x^{2}-1\right)^{\frac{5}{2}}}, x > 1$$ such that <br/><br/>$$y(2)=\frac{2}{9} \log _{e}(2+\sqrt{3}) \text { and } y(\sqrt{2})=\alpha \log _{e}(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}, \alpha, \beta, \gamma \in \mathbb{N} \text {, then } \alpha \beta \gamma \text { is equal to }$$ :</p> | [] | null | 6 | <p>We can solve the given differential equation using an integrating factor.
<br/><br/>The integrating factor is given by :
<br/><br/>$$
\mu(x) = e^{\int \frac{4x}{x^2 - 1} dx} = e^{2\ln(x^2 - 1)} = (x^2 - 1)^2
$$
<br/><br/>Multiplying both sides of the differential equation by $\mu(x)$, we get :</p>
<p>$$(x^2-1)^2 \frac{d y}{d x} + 4x y = \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}}$$</p>
<p>We can rewrite the left-hand side using the product rule:</p>
<p>$$\frac{d}{dx} \left((x^2-1)^2 y\right) = \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}}$$</p>
<p>Integrating both sides with respect to $x$, we get:</p>
<p>$$(x^2-1)^2 y = \int \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}} dx = \sqrt{x^2-1}+2 \ln\left|x+\sqrt{x^2-1}\right|+C$$</p>
<p>where $C$ is the constant of integration. Using the initial condition $y(2) = \frac{2}{9} \log_e(2+\sqrt{3})$, we can solve for $C$:</p>
At $x=2$,
<br/><br/>$9 \cdot \frac{2}{9} \ln (2+\sqrt{3})=2 \ln (2+\sqrt{3})+\sqrt{3}+C$
<br/><br/>$C=-\sqrt{3}$
<br/><br/>At $x=\sqrt{2}$
<br/><br/>$y(\sqrt{2})=2 \ln (1+\sqrt{2})+1-\sqrt{3}$
<br/><br/>$\beta=1, \alpha=2, \gamma=3$
<br/><br/>$$
\Rightarrow \alpha \beta \gamma=6
$$ | integer | jee-main-2023-online-13th-april-evening-shift |
1lgreaalg | maths | differential-equations | linear-differential-equations | <p>Let $$y=y(x), y > 0$$, be a solution curve of the differential equation $$\left(1+x^{2}\right) \mathrm{d} y=y(x-y) \mathrm{d} x$$. If $$y(0)=1$$ and $$y(2 \sqrt{2})=\beta$$, then</p> | [{"identifier": "A", "content": "$$e^{\\beta^{-1}}=e^{-2}(3+2 \\sqrt{2})$$"}, {"identifier": "B", "content": "$$e^{3 \\beta^{-1}}=e(5+\\sqrt{2})$$"}, {"identifier": "C", "content": "$$e^{3 \\beta^{-1}}=e(3+2 \\sqrt{2})$$"}, {"identifier": "D", "content": "$$e^{\\beta^{-1}}=e^{-2}(5+\\sqrt{2})$$"}] | ["C"] | null | $$
\begin{aligned}
& \left(1+x^2\right) d y=y(x-y) d x \\\\
& y(0)=1 \cdot y(2 \sqrt{2})=\beta \\\\
& \frac{d y}{d x}=\frac{y x-y^2}{1+x^2} \\\\
& \frac{d y}{d x}+y\left(\frac{-x}{1+x^2}\right)=\left(\frac{-1}{1+x^2}\right) y^2 \\\\
& \frac{1}{y^2} \frac{d y}{d x}+\frac{1}{y}\left(\frac{-x}{1+x^2}\right)=\frac{-1}{1+x^2} \\\\
& \text { put } \frac{1}{y}=t \text { then } \frac{-1}{y^2} \frac{d y}{d x}=\frac{d t}{d x}
\end{aligned}
$$
<br/><br/>$$
\frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t} \frac{\mathrm{x}}{1+\mathrm{x}^2}=\frac{1}{1+\mathrm{x}^2}
$$
<br/><br/>$$
\text { I.F }=\mathrm{e}^{\int \frac{\mathrm{x}}{1+\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{\frac{1}{2} \ln \left(1+\mathrm{x}^2\right)}=\sqrt{1+\mathrm{x}^2}
$$
<br/><br/>$$
t \sqrt{1+x^2}=\int \frac{\sqrt{1+x^2}}{1+x^2} d x
$$
<br/><br/>$$
\begin{aligned}
& \frac{1}{y} \sqrt{1+x^2}=\int \frac{1}{\sqrt{1+x^2}} d x \\\\
& \frac{1}{y} \sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+C \\\\
& \because y(0)=1 \Rightarrow C=1 \\\\
& \frac{1}{y} \sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+1 \\\\
& \text { For } y=2 \sqrt{2} \\\\
& \frac{3}{y}=\ln |2 \sqrt{2}+3|+1
\end{aligned}
$$
<br/><br/>$$
\begin{gathered}
y=\beta=\frac{3}{1+\ln |2 \sqrt{2}+3|} \\\\
\Rightarrow 3 \beta^{-1}=1+\ln |2 \sqrt{2}+3|
\end{gathered}
$$
<br/><br/>Isolating the term $e^{3 \beta^{-1}}$, we get :
<br/><br/>$$e^{3 \beta^{-1}} = e^{1+\ln |2 \sqrt{2}+3|}.$$
<br/><br/>This can be simplified using the rule $e^{a+b} = e^a e^b$ to :
<br/><br/>$$e^{3 \beta^{-1}} = e \cdot e^{\ln |2 \sqrt{2}+3|}.$$
<br/><br/>Since $e^{\ln(x)} = x$ for any $x$, this simplifies to :
<br/><br/>$$e^{3 \beta^{-1}} = e |2 \sqrt{2}+3|.$$
<br/><br/>Using the given value of $\beta$, which is $\frac{3}{1+\ln |2 \sqrt{2}+3|}$, we find :
<br/><br/>$$e^{3 \beta^{-1}} = e |2 \sqrt{2}+3| = e (2\sqrt{2} + 3),$$
<br/><br/>Since $2\sqrt{2}+3$ is positive and so the absolute value does not affect the result. | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgxw1kmc | maths | differential-equations | linear-differential-equations | <p>Let $$f$$ be a differentiable function such that $${x^2}f(x) - x = 4\int\limits_0^x {tf(t)dt} $$, $$f(1) = {2 \over 3}$$. Then $$18f(3)$$ is equal to :</p> | [{"identifier": "A", "content": "160"}, {"identifier": "B", "content": "210"}, {"identifier": "C", "content": "150"}, {"identifier": "D", "content": "180"}] | ["A"] | null | Given that
<br/><br/>$$
x^2 f(x)-x=4 \int_0^x t f(t) d t
$$
<br/><br/>On differentiating both sides with respect to $x$, we get
<br/><br/>$$
\begin{array}{rlrl}
& 2 x f(x)+x^2 f^{\prime}(x)-1 =4 x f(x) \\\\
&\Rightarrow x^2 f^{\prime}(x)-2 x f(x)-1 =0 \\\\
&\Rightarrow x^2 \frac{d y}{d x}-2 x y =1 ~~~~~~~(Let, y=f(x) ]\\\\
&\frac{d y}{d x}-\frac{2}{x} y =\frac{1}{x^2}
\end{array}
$$
<br/><br/>On comparing above equation with
<br/><br/>$\frac{d y}{d x}+P y=Q$, where $P=-\frac{2}{x}, Q=\frac{1}{x^2}$
<br/><br/>Now, IF $=e^{\int(-2 / x) d x}=e^{-2 \log x}=1 / x^2$
<br/><br/>Solution is $y\left(\frac{1}{x^2}\right)=\int\left(\frac{1}{x^2}\right) \frac{1}{x^2} d x$
<br/><br/>$$
\begin{array}{ll}
&\Rightarrow \frac{y}{x^2}=\int \frac{1}{x^4} d x \Rightarrow \frac{y}{x^2}=-\frac{1}{3 x^3}+C \\\\
&\Rightarrow y =-\frac{1}{3 x}+C x^2
\end{array}
$$
<br/><br/>Given, $f(1)=\frac{2}{3}$.
<br/><br/>So, $\frac{2}{3}=-\frac{1}{3}+C \Rightarrow C=1$
<br/><br/>Thus, $y=f(x)=-\frac{1}{3 x}+x^2$
<br/><br/>$$
\therefore 18 f(3)=18\left\{9-\frac{1}{9}\right\}=18 \times \frac{80}{9}=160
$$ | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgyq0yd0 | maths | differential-equations | linear-differential-equations | <p>Let the solution curve $$x=x(y), 0 < y < \frac{\pi}{2}$$, of the differential equation $$\left(\log _{e}(\cos y)\right)^{2} \cos y \mathrm{~d} x-\left(1+3 x \log _{e}(\cos y)\right) \sin \mathrm{y} d y=0$$ satisfy $$x\left(\frac{\pi}{3}\right)=\frac{1}{2 \log _{e} 2}$$. If $$x\left(\frac{\pi}{6}\right)=\frac{1}{\log _{e} m-\log _{e} n}$$, where $$m$$ and $$n$$ are coprime, then $$m n$$ is equal to __________.</p> | [] | null | 12 | $$
\begin{aligned}
& (\cos y)(\ln (\cos y))^2 d x=(1+3 x \ln \cos y) \sin y d y \\\\
& \Rightarrow \frac{d x}{d y}=\frac{(1+3 x \ln \cos y) \sin y}{(\ln \cos y)^2 \cos y} \\\\
& =\tan y\left[\frac{1}{(\ln \cos y)^2}+\frac{3 x}{\ln \cos y}\right] \\\\
& \Rightarrow \frac{d x}{d y}-\left(\frac{3 \tan y}{\ln \cos y}\right) x=\frac{\tan y}{(\ln \cos y)^2}
\end{aligned}
$$
<br/><br/>Which is a linear differential equation.
<br/><br/>$$
\text { I.F. }=e^{-\int \frac{3 \tan y}{\ln \cos y} d y}=(\ln \cos y)^3
$$
<br/><br/>So, the solution is :
<br/><br/>$$
\begin{aligned}
& x \times(\ln \cos y)^3=\int\left((\ln \cos y)^3 \times \frac{\tan y}{(\ln \cos y)^2}\right) d y \\\\
& x \times(\ln \cos y)^3=\frac{-(\ln \cos y)^2}{2}+C
\end{aligned}
$$
<br/><br/>$$
\text { At } y=\frac{\pi}{3} \text {, }
$$
<br/><br/>$$
\begin{aligned}
& \frac{1}{2 \ln 2} \times\left(\ln \left(\frac{1}{2}\right)\right)^3=-\frac{\left(\ln \left(\frac{1}{2}\right)\right)^2}{2}+C \\\\
& \Rightarrow C=0
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { So, } x \times \ln ^3 \cos y=\frac{-\ln ^2 \cos y}{2} \\\\
& \text { At } y=\frac{\pi}{6}, x \times\left(\ln \left(\frac{\sqrt{3}}{2}\right)\right)^3=-\frac{1}{2}\left(\ln \left(\frac{\sqrt{3}}{2}\right)\right)^2 \\\\
& \Rightarrow x=-\frac{1}{2 \ln \left(\frac{\sqrt{3}}{2}\right)}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =-\frac{1}{2[\ln \sqrt{3}-\ln 2]}=\frac{-1}{2\left[\frac{1}{2} \ln 3-\ln 2\right]} \\\\
& =\frac{-1}{2\left[\frac{\ln 3-\ln 4}{2}\right]}=\frac{1}{\ln 4-\ln 3} \\\\
& \Rightarrow m=4, n=3 \\\\
& \Rightarrow m n=12
\end{aligned}
$$ | integer | jee-main-2023-online-8th-april-evening-shift |
1lh00e9y0 | maths | differential-equations | linear-differential-equations | <p>If the solution curve of the differential equation $$\left(y-2 \log _{e} x\right) d x+\left(x \log _{e} x^{2}\right) d y=0, x > 1$$ passes through the points $$\left(e, \frac{4}{3}\right)$$ and $$\left(e^{4}, \alpha\right)$$, then $$\alpha$$ is equal to ____________.</p> | [] | null | 3 | The given differential equation is,
<br/><br/>$$
\begin{aligned}
& (y-2 \log x) d x+\left(x \log x^2\right) d y=0 \\\\
& \Rightarrow \frac{d y}{d x}=\frac{(2 \log x-y)}{2 x \log x} \\\\
& \Rightarrow \frac{d y}{d x}+\frac{y}{2 x \log x}=\frac{1}{x}
\end{aligned}
$$
<br/><br/>It is a linear differential equation.
<br/><br/>$$
\therefore \text { I.F. }=e^{\int \frac{1}{2 x \log x} d x}
$$
<br/><br/>Put $\log x=t \Rightarrow \frac{1}{x} d x=d t$
<br/><br/>$$
\therefore \text { I.F. }=e^{\int \frac{1}{2 t} d t}=e^{\log (t)^{\frac{1}{2}}}=\sqrt{t}=\sqrt{\log x}
$$
<br/><br/>So, required solution is,
<br/><br/>$$
\begin{aligned}
& y \sqrt{\log x}=\int \frac{\sqrt{\log x}}{x} d x \\\\
& \log x=v \Rightarrow \frac{1}{x} d x=d v \\\\
& \Rightarrow y \sqrt{\log x}=\int \sqrt{v} d v+C \\\\
& \Rightarrow y \sqrt{\log x}=\frac{2 v^{3 / 2}}{3}+C \\\\
& \Rightarrow y \sqrt{\log x}=\frac{2}{3}(\log x)^{3 / 2}+C
\end{aligned}
$$
<br/><br/>Now, this curve passes through $\left(e, \frac{4}{3}\right)$ and $\left(e^4, \alpha\right)$
<br/><br/>$$
\begin{aligned}
& \therefore \frac{4}{3} \sqrt{\log e}=\frac{2}{3}(\log e)^{3 / 2}+C \\\\
& \Rightarrow C=\frac{4}{3}-\frac{2}{3}=\frac{2}{3}
\end{aligned}
$$
<br/><br/>Also, $\alpha \sqrt{\log e^4}=\frac{2}{3}\left(\log e^4\right)^{3 / 2}+\frac{2}{3}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow 2 \alpha=\frac{2}{3} \times(4)^{3 / 2}+\frac{2}{3}=\frac{16}{3}+\frac{2}{3}=\frac{18}{3} \\\\
& \Rightarrow \alpha=3
\end{aligned}
$$ | integer | jee-main-2023-online-8th-april-morning-shift |
1lh23pwxa | maths | differential-equations | linear-differential-equations | <p>Let $$y=y(x)$$ be a solution of the differential equation $$(x \cos x) d y+(x y \sin x+y \cos x-1) d x=0,0 < x < \frac{\pi}{2}$$. If $$\frac{\pi}{3} y\left(\frac{\pi}{3}\right)=\sqrt{3}$$, then $$\left|\frac{\pi}{6} y^{\prime \prime}\left(\frac{\pi}{6}\right)+2 y^{\prime}\left(\frac{\pi}{6}\right)\right|$$ is equal to ____________.</p> | [] | null | 2 | Given, differential equation
<br/><br/>$$
\begin{aligned}
& (x \cos x) d y+(x y \sin x+y \cos x-1) d x=0,0 < x < \frac{\pi}{2} \\\\
& \Rightarrow \frac{d y}{d x}+\frac{x y \sin x+y \cos x-1}{x \cos x}=0 \\\\
& \Rightarrow \frac{d y}{d x}+\left(\frac{x y \sin x+y \cos x}{x \cos x}\right)=\frac{1}{x \cos x} \\\\
& \Rightarrow \frac{d y}{d x}+\left(\tan x+\frac{1}{x}\right) y=\frac{1}{x \cos x}
\end{aligned}
$$
<br/><br/>Which is linear differential equation in the form of
<br/><br/>$$
\begin{aligned}
& \frac{d y}{d x}+P y=Q \\\\
& \therefore \mathrm{IF}=e^{\int\left(\tan x+\frac{1}{x}\right) d x}=e^{(\log \sec x+\log x)}=e^{\log (x \sec x)}=x \sec x
\end{aligned}
$$
<br/><br/>$\therefore$ The general solution of the given differential equation
<br/><br/>$$
\begin{array}{rlrl}
y \cdot \mathrm{IF} =\int(Q \times \mathrm{IF}) d x+c \\\\
\Rightarrow y(x \sec x) =\int\left(\frac{1}{x \cos x} x \sec x\right) d x+c \\\\
\Rightarrow x y \sec x =\int \sec ^2 x d x+c \\\\
\Rightarrow x y \sec x =\tan x+c ........(i)
\end{array}
$$
<br/><br/>Since, $\frac{\pi}{3} y\left(\frac{\pi}{3}\right)=\sqrt{3} \Rightarrow y\left(\frac{\pi}{3}\right)=\frac{3 \sqrt{3}}{\pi}$
<br/><br/>$$
\begin{aligned}
& \therefore \frac{\pi}{3}\left(\frac{3 \sqrt{3}}{\pi}\right) \sec \left(\frac{\pi}{3}\right)=\tan \left(\frac{\pi}{3}\right)+c \\\\
& \Rightarrow \sqrt{3}(2)=\sqrt{3}+c \\\\
& \begin{aligned}
\Rightarrow & c=\sqrt{3}
\end{aligned}
\end{aligned}
$$
<br/><br/>On putting the value of $c$ in Eq. (i), we get
<br/><br/>$$
\begin{aligned}
&x y \sec x =\tan x+\sqrt{3} \\\\
&\Rightarrow y =\frac{1}{x}(\sin x+\sqrt{3} \cos x) \\\\
& y^{\prime} =\frac{1}{x}(\cos x-\sqrt{3} \sin x)-\frac{1}{x^2}(\sin x+\sqrt{3} \cos x)
\end{aligned}
$$
<br/><br/>$\begin{aligned} & \text { and } y^{\prime \prime}=\frac{1}{x}(-\sin x-\sqrt{3} \cos x)-\frac{1}{x^2}(\cos x-\sqrt{3} \sin x) -\frac{1}{x^2}(\cos x-\sqrt{3} \sin x) \\ & -\frac{2}{x^3}(\cos x-\sqrt{3} \sin x) \\\\ & \therefore\left|\frac{\pi}{6} y^{\prime \prime}\left(\frac{\pi}{6}\right)+2 y^{\prime}\left(\frac{\pi}{6}\right)\right|=|-2|=2 \\\\ & \end{aligned}$ | integer | jee-main-2023-online-6th-april-morning-shift |
lsamlzre | maths | differential-equations | linear-differential-equations | Let $\alpha$ be a non-zero real number. Suppose $f: \mathbf{R} \rightarrow \mathbf{R}$ is a differentiable function such that $f(0)=2$ and $\lim\limits_{x \rightarrow-\infty} f(x)=1$. If $f^{\prime}(x)=\alpha f(x)+3$, for all $x \in \mathbf{R}$, then $f\left(-\log _{\mathrm{e}} 2\right)$ is equal to : | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "5"}] | ["B"] | null | $$
\begin{aligned}
& f(0)=2, \lim _{x \rightarrow-\infty} f(x)=1 \\\\
& f^{\prime}(x)-x \cdot f(x)=3 \\\\
& \text { I.F }=e^{-\alpha x} \\\\
& y\left(e^{-\alpha x}\right)=\int 3 \cdot e^{-\alpha x} d x \\\\
& f(x) \cdot\left(e^{-\alpha x}\right)=\frac{3 e^{-\alpha x}}{-\alpha}+c \\\\
& x=0 \Rightarrow 2=\frac{-3}{\alpha}+c \Rightarrow \frac{3}{\alpha}=c-2 \\\\
& f(x)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\
& x \rightarrow-\infty \Rightarrow 1=\frac{-3}{\alpha}+c(0) \\\\
& \alpha=-3 \therefore c=1 \\\\
& f(-\ln 2)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\
& =1+e^{3 \ln 2}=9
\end{aligned}
$$
| mcq | jee-main-2024-online-1st-february-evening-shift |
lsapuxwg | maths | differential-equations | linear-differential-equations | If $x=x(t)$ is the solution of the differential equation $(t+1) \mathrm{d} x=\left(2 x+(t+1)^4\right) \mathrm{dt}, x(0)=2$, then, $x(1)$ equals _________. | [] | null | 14 | $\begin{aligned} & (\mathrm{t}+1) \mathrm{dx}=\left(2 \mathrm{x}+(\mathrm{t}+1)^4\right) \mathrm{dt} \\\\ & \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2 \mathrm{x}+(\mathrm{t}+1)^4}{\mathrm{t}+1} \\\\ & \frac{\mathrm{dx}}{\mathrm{dt}}-\frac{2 \mathrm{x}}{\mathrm{t}+1}=(\mathrm{t}+1)^3\end{aligned}$
<br/><br/>$I.F=e^{-\int \frac{2}{t+1} d t}=e^{-2 \ln (t+1)}=\frac{1}{(t+1)^2}$
<br/><br/>$\begin{aligned} & \frac{x}{(t+1)^2}=\int \frac{1}{(t+1)^2}(t+1)^3 d t+C\end{aligned}$
<br/><br/>$\frac{x}{(t+1)^2}=\frac{t^2}{2}+t+C$
<br/><br/>Now $x(0)=2$
<br/><br/>$$
\begin{aligned}
& \Rightarrow C=2 \\\\
& \therefore x=\left(\frac{t^2}{2}+t+2\right)(t+1)^2 \\\\
& x(1)=\left(\frac{1}{2}+1+2\right)(1+1)^2 \\\\
& =\frac{7}{2} \times 4=14
\end{aligned}
$$ | integer | jee-main-2024-online-1st-february-morning-shift |
lsbkktpn | maths | differential-equations | linear-differential-equations | Let $x=x(\mathrm{t})$ and $y=y(\mathrm{t})$ be solutions of the differential equations $\frac{\mathrm{d} x}{\mathrm{dt}}+\mathrm{a} x=0$ and $\frac{\mathrm{d} y}{\mathrm{dt}}+\mathrm{by}=0$ respectively, $\mathrm{a}, \mathrm{b} \in \mathbf{R}$. Given that $x(0)=2 ; y(0)=1$ and $3 y(1)=2 x(1)$, the value of $\mathrm{t}$, for which $x(\mathrm{t})=y(\mathrm{t})$, is : | [{"identifier": "A", "content": "$\\log _{\\frac{2}{3}} 2$"}, {"identifier": "B", "content": "$\\log _{\\frac{4}{3}} 2$"}, {"identifier": "C", "content": "$\\log _4 3$"}, {"identifier": "D", "content": "$\\log _3 4$"}] | ["B"] | null | <p>$$\begin{aligned}
& \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0 \\
& \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{adt} \\
& \int \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{a} \int \mathrm{dt} \\
& \ln |\mathrm{x}|=-\mathrm{at}+\mathrm{c} \\
& \mathrm{at} \mathrm{t}=0, \mathrm{x}=2 \\
& \ln 2=0+\mathrm{c} \\
& \ln \mathrm{x}=-\mathrm{at}+\ln 2 \\
& \frac{\mathrm{x}}{2}=\mathrm{e}^{-\mathrm{at}} \\
& \mathrm{x}=2 \mathrm{e}^{-\mathrm{at}} \quad \text{.... (i)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{d y}{d t}+b y=0 \\
& \frac{d y}{y}=-b d t \\
& \ln |y|=-b t+\lambda \\
& t=0, y=1 \\
& 0=0+\lambda \\
& y=e^{-b t} \quad \text{..... (ii)}
\end{aligned}$$</p>
<p>According to question</p>
<p>$$\begin{aligned}
& 3 \mathrm{y}(1)=2 \mathrm{x}(1) \\
& 3 \mathrm{e}^{-\mathrm{b}}=2\left(2 \mathrm{e}^{-\mathrm{a}}\right) \\
& \mathrm{e}^{\mathrm{a}-\mathrm{b}}=\frac{4}{3}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { For } x(t)=y(t) \\
& \begin{array}{r}
2 \mathrm{e}^{-a t}=e^{-b t} \\
2=e^{(a-b) t} \\
2=\left(\frac{4}{3}\right)^t \\
\log _{\frac{4}{3}} 2=t
\end{array}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lsd4l5na | maths | differential-equations | linear-differential-equations | <p>The temperature $$T(t)$$ of a body at time $$t=0$$ is $$160^{\circ} \mathrm{F}$$ and it decreases continuously as per the differential equation $$\frac{d T}{d t}=-K(T-80)$$, where $$K$$ is a positive constant. If $$T(15)=120^{\circ} \mathrm{F}$$, then $$T(45)$$ is equal to</p> | [{"identifier": "A", "content": "90$$^\\circ$$ F"}, {"identifier": "B", "content": "85$$^\\circ$$ F"}, {"identifier": "C", "content": "80$$^\\circ$$ F"}, {"identifier": "D", "content": "95$$^\\circ$$ F"}] | ["A"] | null | <p>$$\begin{aligned}
& \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-80) \\
& \int_\limits{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int_\limits0^{\mathrm{t}}-\mathrm{Kdt} \\
& {[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}} \\
& \ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt} \\
& \ln \left|\frac{\mathrm{T}-80}{80}\right|=-\mathrm{kt} \\
& \mathrm{T}=80+80 \mathrm{e}^{-\mathrm{kt}} \\
& 120=80+80 \mathrm{e}^{-\mathrm{k} .15} \\
& \frac{40}{80}=\mathrm{e}^{-\mathrm{k} 15}=\frac{1}{2} \\
& \therefore \mathrm{T}(45)=80+80 \mathrm{e}^{-\mathrm{k} .45} \\
& =80+80\left(\mathrm{e}^{-\mathrm{k} .15}\right)^3 \\
& =80+80 \times \frac{1}{8} \\
& =90
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lsf0ewla | maths | differential-equations | linear-differential-equations | <p>A function $$y=f(x)$$ satisfies $$f(x) \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0$$ with condition $$f(0)=0$$. Then, $$f\left(\frac{\pi}{2}\right)$$ is equal to</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "0"}] | ["B"] | null | <p>$$\begin{aligned}
& \frac{d y}{d x}-\left(\frac{\sin 2 x}{1+\cos ^2 x}\right) y=\sin x \\
& \text { I.F. }=1+\cos ^2 x \\
& y \cdot\left(1+\cos ^2 x\right)=\int(\sin x) d x \\
& =-\cos x+C \\
& x=0, C=1 \\
& y\left(\frac{\pi}{2}\right)=1
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
1lsgb4t6h | maths | differential-equations | linear-differential-equations | <p>Let $$y=y(x)$$ be the solution of the differential equation $$\left(1-x^2\right) \mathrm{d} y=\left[x y+\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}\right] \mathrm{d} x, -1< x<1, y(0)=0$$. If $$y\left(\frac{1}{2}\right)=\frac{\mathrm{m}}{\mathrm{n}}, \mathrm{m}$$ and $$\mathrm{n}$$ are co-prime numbers, then $$\mathrm{m}+\mathrm{n}$$ is equal to __________.</p> | [] | null | 97 | <p>$$\begin{aligned}
& \frac{d y}{d x}-\frac{x y}{1-x^2}=\frac{\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}}{1-x^2} \\
& \mathrm{IF}=e^{-\int \frac{x}{1-x^2} d x}=e^{+\frac{1}{2} \ln \left(1-x^2\right)}=\sqrt{1-x^2} \\
& y \sqrt{1-x^2}=\sqrt{3} \int\left(x^3+2\right) d x \\
& y \sqrt{1-x^2}=\sqrt{3}\left(\frac{x^4}{4}+2 x\right)+c \\
& \Rightarrow y(0)=0 \quad \therefore c=0\\
& y\left(\frac{1}{2}\right)=\frac{65}{32}=\frac{m}{n} \\
& m+n=97
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-morning-shift |
luxweayj | maths | differential-equations | linear-differential-equations | <p>Let $$\int_\limits0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t, 0 \leq x \leq 3, y \geq 0, y(0)=0$$. Then at $$x=2, y^{\prime \prime}+y+1$$ is equal to</p> | [{"identifier": "A", "content": "$$\\sqrt2$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1/2"}, {"identifier": "D", "content": "1"}] | ["D"] | null | <p>$$\int_\limits0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_\limits0^x y(t) d t$$</p>
<p>Differentiating both side</p>
<p>$$\begin{aligned}
& \sqrt{1-\left(y^{\prime}(x)\right)^2}=y(x) \\
& \left(\frac{d y}{d x}\right)^2+y^2=1 \\
& y^{\prime 2}+y^2=1 \\
& 2 y^{\prime} y^{\prime \prime}+2 y y^{\prime}=0 \\
& y^{\prime \prime}+y=0
\end{aligned}$$</p>
<p>$$\therefore$$ <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw1mj9xg/0016d131-6a79-41e3-a383-e930f3bafac3/739e0c40-0f51-11ef-91cd-f19f7dc20f18/file-1lw1mj9xh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw1mj9xg/0016d131-6a79-41e3-a383-e930f3bafac3/739e0c40-0f51-11ef-91cd-f19f7dc20f18/file-1lw1mj9xh.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Evening Shift Mathematics - Differential Equations Question 19 English Explanation"></p> | mcq | jee-main-2024-online-9th-april-evening-shift |
luxweoxi | maths | differential-equations | linear-differential-equations | <p>For a differentiable function $$f: \mathbb{R} \rightarrow \mathbb{R}$$, suppose $$f^{\prime}(x)=3 f(x)+\alpha$$, where $$\alpha \in \mathbb{R}, f(0)=1$$ and $$\lim _\limits{x \rightarrow-\infty} f(x)=7$$. Then $$9 f\left(-\log _e 3\right)$$ is equal to _________.</p> | [] | null | 61 | <p>$$\begin{aligned}
& f^{\prime}(x)=3 f(x)+\alpha \\
& \Rightarrow \frac{d y}{3 y+\alpha}=d x \\
& \Rightarrow \frac{1}{3} \ln (3 y+\alpha)=x+C \\
& y(0)=1 \Rightarrow C=\frac{1}{3} \ln (3+\alpha) \\
& \frac{1}{3} \ln \left(\frac{3 y+\alpha}{3+\alpha}\right)=x \\
& \Rightarrow y=\frac{1}{3}\left((3+\alpha) e^{3 x}-\alpha\right)=f(x) \\
& \lim _{x \rightarrow-\infty} f(x)=7 \Rightarrow \alpha=-21 \\
& \Rightarrow f(x)=7-6 e^{3 x} \\
& 9 f(-\ln 3)=61
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-evening-shift |
luy6z4dh | maths | differential-equations | linear-differential-equations | <p>The solution of the differential equation $$(x^2+y^2) \mathrm{d} x-5 x y \mathrm{~d} y=0, y(1)=0$$, is :</p> | [{"identifier": "A", "content": "$$\\left|x^2-4 y^2\\right|^5=x^2$$\n"}, {"identifier": "B", "content": "$$\\left|x^2-2 y^2\\right|^6=x$$\n"}, {"identifier": "C", "content": "$$\\left|x^2-2 y^2\\right|^5=x^2$$\n"}, {"identifier": "D", "content": "$$\\left|x^2-4 y^2\\right|^6=x$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \left(x^2+y^2\right) d x-5 x y d y=0 \\
& \frac{d y}{d x}=\frac{x^2+y^2}{5 x y} \\
& \text { Let } y=v x \\
& \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& V+x \frac{d v}{d x}=\frac{1+v^2}{5 v} \\
& x \frac{d v}{d x}=\frac{1+v^2-5 v^2}{5 v}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{1}{8} \int \frac{8 \times 5 v d v}{1-4 v^2}=\int \frac{d x}{x} \\
& \frac{-5}{8} \ln \left|1-4 v^2\right|=\ln |x|+\ln c \\
& \Rightarrow \frac{5}{8} \ln \frac{\left|x^2-4 y^2\right|}{x^2}+\ln |x|=\ln c \\
& \left|\frac{x^2-4 y^2}{x^2}\right|^{5 / 8}|x|=c \\
& \frac{\left|x^2-4 y^2\right|^{5 / 8}}{|x|^{\frac{1}{4}}}=c \because y(1)=0 \Rightarrow c=1 \\
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow\left|x^2-4 y^2\right|^{5 / 8}=|x|^{\frac{1}{4}} \\
& \left|x^2-4 y^2\right|^5=x^2
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
luy6z5jr | maths | differential-equations | linear-differential-equations | <p>The solution curve, of the differential equation $$2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+3=5 \frac{\mathrm{d} y}{\mathrm{~d} x}$$, passing through the point $$(0,1)$$ is a conic, whose vertex lies on the line :</p> | [{"identifier": "A", "content": "$$2 x+3 y=-9$$\n"}, {"identifier": "B", "content": "$$2 x+3 y=-6$$\n"}, {"identifier": "C", "content": "$$2 x+3 y=9$$\n"}, {"identifier": "D", "content": "$$2 x+3 y=6$$"}] | ["C"] | null | <p>$$\begin{aligned}
& 2 y \frac{d y}{d x}+3=5 \frac{d y}{d x} \\
& 2 y d y+3 d x=5 d y \\
& y^2+3 x=5 y+\left.c\right|_{(0,1)} \\
& 1+0=5+c \\
& c=-4 \\
& y^2-5 y=-3 x-4 \\
& y^2-5 y+\frac{25}{4}-\frac{25}{4}=-3 x-4 \\
& \left(y-\frac{5}{2}\right)^2=-3 x+\frac{9}{4} \\
& \left(y-\frac{5}{2}\right)^2=-3\left(x-\frac{3}{4}\right) \\
& \left(\frac{3}{4}, \frac{5}{2}\right) \text { satisfies by } 2 x+3 y=9
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
lv0vxdf0 | maths | differential-equations | linear-differential-equations | <p>Let the solution $$y=y(x)$$ of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}-y=1+4 \sin x$$ satisfy $$y(\pi)=1$$. Then $$y\left(\frac{\pi}{2}\right)+10$$ is equal to __________.</p> | [] | null | 7 | <p>$$\begin{aligned}
& \frac{d y}{d x}-y=1+4 \sin x \\
& \text { Integrating factor }=e^{-\int d x}=e^{-x}
\end{aligned}$$</p>
<p>Solution is $$y e^{-x}=\int(1+4 \sin x) e^{-x} d x$$</p>
<p>$$\begin{aligned}
& =-e^{-x}+2 \cdot e^{-x}(-\sin x-\cos x)+C \\
y(\pi) & =1 \Rightarrow C=0
\end{aligned}$$</p>
<p>Hence $$y(x)=-1-2(\sin x+\cos x)$$</p>
<p>$$y\left(\frac{\pi}{2}\right)+10=7$$</p> | integer | jee-main-2024-online-4th-april-morning-shift |
lv7v4l5j | maths | differential-equations | linear-differential-equations | <p>If $$y=y(x)$$ is the solution of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=\sin (2 x), y(0)=\frac{3}{4}$$, then $$y\left(\frac{\pi}{8}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\mathrm{e}^{-\\pi / 8}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{e}^{\\pi / 4}$$\n"}, {"identifier": "C", "content": "$$\\mathrm{e}^{-\\pi / 4}$$\n"}, {"identifier": "D", "content": "$$\\mathrm{e}^{\\pi / 8}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \frac{d y}{d x}+2 y=\sin 2 x \\
& \text { IF }=e^{2 d x}=e^{2 x} \\
& y \cdot e^{2 x}=\int e^{2 x} \sin 2 x d x+c \\
& =\frac{e^{2 x}}{8}(2 \sin 2 x-2 \cos 2 x)+c \\
& y(0)=\frac{3}{4} \\
& \frac{3}{4}=\frac{1}{8}(-2)+c \Rightarrow c=1 \\
& \text { Put } x=\frac{\pi}{8} \\
& y=\frac{1}{8} \times 2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)+e^{-\pi / 4} \\
& y=e^{-\pi / 4}
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
lv9s20im | maths | differential-equations | linear-differential-equations | <p>Let $$y=y(x)$$ be the solution of the differential equation</p>
<p>$$\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 x}{\left(1+x^2\right)^2} y=x \mathrm{e}^{\frac{1}{\left(1+x^2\right)}} ; y(0)=0.$$</p>
<p>Then the area enclosed by the curve $$f(x)=y(x) \mathrm{e}^{-\frac{1}{\left(1+x^2\right)}}$$ and the line $$y-x=4$$ is ________.</p> | [] | null | 18 | <p>$$\frac{d y}{d x}+\frac{2 x}{\left(1+x^2\right)^2} y=x e^{\frac{1}{1+x^2}} ; y(0)=0$$</p>
<p>I.F. of linear differential equation,</p>
<p>$$\begin{aligned}
& \text { I.F. }=e^{\int \frac{2 x}{\left(1+x^2\right)^2}} d x=e^{\left(\frac{-1}{1+x^2}\right)} \\
& \Rightarrow y\left(e^{\left(\frac{-1}{1+x^2}\right)}\right)=\int x \cdot e^{\frac{1}{1+x^2}} \cdot e^{\left(\frac{-1}{1+x^2}\right)} d x \\
& =\frac{x^2}{2}+c \\
& \Rightarrow y(0)=0 \Rightarrow 0\left(e^{-1}\right)=c \Rightarrow c=0 \\
& \Rightarrow y=\frac{e^{\frac{1}{1+x^2}} \cdot x^2}{2}
\end{aligned}$$</p>
<p>Area between curve $$y e^{\left(\frac{-1}{1+x^2}\right)}=\frac{x^2}{2}$$ and $$y-x=4$$</p>
<p>$$\begin{aligned}
& \Rightarrow 2(x+4)=x^2 \Rightarrow x^2-2 x-8=0 \\
& \Rightarrow(x-4)(x+2)=0 \\
& \int_{-2}^4\left[(x+4)-\frac{x^2}{2}\right] d x=\frac{x^2}{2}+4 x-\left.\frac{x^3}{6}\right|_{-2} ^4 \\
& =\left(8+16-\frac{64}{6}\right)-\left(2-8+\frac{8}{6}\right) \\
& =30-12=18
\end{aligned}$$</p> | integer | jee-main-2024-online-5th-april-evening-shift |
lvc57baa | maths | differential-equations | linear-differential-equations | <p>Let $$y=y(x)$$ be the solution of the differential equation $$\left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x>0$$ and $$y\left(e^{-1}\right)=0$$. Then, $$y(e)$$ is equal to </p> | [{"identifier": "A", "content": "$$-\\frac{3}{\\mathrm{e}}$$\n"}, {"identifier": "B", "content": "$$-\\frac{3}{2 \\mathrm{e}}$$\n"}, {"identifier": "C", "content": "$$-\\frac{2}{3 \\mathrm{e}}$$\n"}, {"identifier": "D", "content": "$$-\\frac{2}{\\mathrm{e}}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& (2 x \log x) \frac{d y}{d x}+2 y=\frac{3}{x} \log x \\
& \frac{d y}{d x}+\frac{y}{x \log x}=\frac{3}{2 x^2} \\
& I F=e^{\int \frac{1}{x \log x}}=e^{\log (\log x)}=\log x \\
& y \times \log x=\int \log x \times \frac{3}{2 x^2} d x+C \\
& y \log x=\frac{3}{2}\left[\frac{-\log x}{x}-\frac{1}{x}\right]+C \\
& y\left(e^{-1}\right)=0 \\
& \Rightarrow 0=\frac{3}{2}[e-e]+C \\
& C=0 \\
& y \log x=\frac{-3}{2}\left[\frac{\log x+1}{x}\right] \\
& y(e) \rightarrow y(e) \times 1=\frac{-3}{2}\left[\frac{1+1}{e}\right]=-\frac{3}{e}
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
lvc58e2y | maths | differential-equations | linear-differential-equations | <p>Let $$y=y(x)$$ be the solution of the differential equation $$\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}$$, $$y(1)=0$$. Then $$y(0)$$ is</p> | [{"identifier": "A", "content": "$$\\frac{1}{4}\\left(e^{\\pi / 2}-1\\right)$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{2}\\left(1-e^{\\pi / 2}\\right)$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{4}\\left(1-e^{\\pi / 2}\\right)$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{2}\\left(e^{\\pi / 2}-1\\right)$$"}] | ["B"] | null | <p>To determine $ y(0) $, we start by solving the differential equation given:</p>
<p>$ (1 + x^2) \frac{dy}{dx} + y = e^{\tan^{-1} x} $</p>
<p>First, we rewrite it in the standard form for a linear differential equation:</p>
<p>$ \frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{e^{\tan^{-1} x}}{1 + x^2} $</p>
<p>Next, we find the integrating factor (I.F.):</p>
<p>$ \text{I.F.} = e^{\int \frac{1}{1 + x^2} dx} = e^{\tan^{-1} x} $</p>
<p>Multiply through by the integrating factor:</p>
<p>$ y \cdot e^{\tan^{-1} x} = \int e^{\tan^{-1} x} \cdot \frac{e^{\tan^{-1} x}}{1 + x^2} dx $</p>
<p>This simplifies to:</p>
<p>$ y \cdot e^{\tan^{-1} x} = \int \frac{e^{2 \tan^{-1} x}}{1 + x^2} dx $</p>
<p>Make the substitution $ \tan^{-1} x = t $, then $ \frac{1}{1 + x^2} dx = dt $:</p>
<p>$ \int e^{2t} dt = \frac{e^{2t}}{2} + \frac{C}{2} $</p>
<p>Rewrite in terms of $ x $:</p>
<p>$ y \cdot e^{\tan^{-1} x} = \frac{e^{2 \tan^{-1} x}}{2} + \frac{C}{2} $</p>
<p>Use the initial condition $ y(1) = 0 $:</p>
<p>$ 0 = \frac{e^{2 \cdot \tan^{-1} 1}}{2} + \frac{C}{2} $</p>
<p>Since $ \tan^{-1} 1 = \frac{\pi}{4} $:</p>
<p>$ 0 = \frac{e^{\pi/2}}{2} + \frac{C}{2} \implies C = -e^{\pi/2} $</p>
<p>Thus, the solution is:</p>
<p>$ y \cdot e^{\tan^{-1} x} = \frac{e^{2 \tan^{-1} x} - e^{\pi/2}}{2} $</p>
<p>Evaluating $ y(0) $:</p>
<p>$ y(0) = y \cdot 1 = \frac{1 - e^{\pi/2}}{2} $</p>
<p>Therefore,</p>
<p>$ y(0) = \frac{1}{2} (1 - e^{\pi/2}) $</p>
<p>Hence, the correct answer is:</p>
<p>Option B</p>
<p>$ \frac{1}{2}\left(1-e^{\pi / 2}\right) $</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
v1C52TlUog2JpJ03 | maths | differential-equations | order-and-degree | The order and degree of the differential equation
<br/>$$\,{\left( {1 + 3{{dy} \over {dx}}} \right)^{2/3}} = 4{{{d^3}y} \over {d{x^3}}}$$ are | [{"identifier": "A", "content": "$$\\left( {1,{2 \\over 3}} \\right)$$ "}, {"identifier": "B", "content": "$$(3, 1)$$ "}, {"identifier": "C", "content": "$$(3,3)$$ "}, {"identifier": "D", "content": "$$(1,2)$$"}] | ["C"] | null | $${\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} =$$ (4)<sup>3</sup>$${\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}$$
<br><br>$$ \Rightarrow {\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} = 16{\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}$$
<br><br>$$ \therefore $$ Order = 3 and Degree = 3 | mcq | aieee-2002 |
74wvDLmkOoJYDYJv | maths | differential-equations | order-and-degree | The degree and order of the differential equation of the family of all parabolas whose axis is $$x$$-axis, are respectively. | [{"identifier": "A", "content": "$$2, 3$$ "}, {"identifier": "B", "content": "$$2,1$$ "}, {"identifier": "C", "content": "$$1,2$$ "}, {"identifier": "D", "content": "$$3,2.$$"}] | ["C"] | null | General equation of parabola whose axis is along X-axis
<br><br>$${y^2} = 4a\left( {x - h} \right)$$
<br><br>Differentiating with respect to x, we get
<br><br>$$2y{y_1} = 4a$$
<br><br>$$ \Rightarrow y{y_1} = 2a$$
<br><br>Again differentiating with respect to x, we get
<br><br>$$ \Rightarrow {y_1}^2 + y{y_2} = 0$$
<br><br>Degree $$=1,$$ order $$=2.$$ | mcq | aieee-2003 |
9wlLHTpefi6qCARx | maths | differential-equations | order-and-degree | The differential equation representing the family of curves $${y^2} = 2c\left( {x + \sqrt c } \right),$$ where $$c>0,$$ is a parameter, is of order and degree as follows: | [{"identifier": "A", "content": "order $$1,$$ degree $$2$$"}, {"identifier": "B", "content": "order $$1,$$ degree $$1$$"}, {"identifier": "C", "content": "order $$1,$$ degree $$3$$"}, {"identifier": "D", "content": "order $$2,$$ degree $$2$$"}] | ["C"] | null | $${y^2} = 2c\left( {x + \sqrt c } \right)\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$2yy' = 2c.1\,\,\,or\,\,\,yy' = c\,\,\,...\left( {ii} \right)$$
<br><br>$$ \Rightarrow {y^2} = 2yy'\left( {x + \sqrt {yy'} } \right)$$
<br><br>$$\left[ \, \right.$$ On putting value of $$c$$ from $$(ii)$$ in $$(i)$$ $$\left. \, \right]$$
<br><br>On simplifying, we get
<br><br>$${\left( {y - 2xy'} \right)^2} = 4yy{'^3}\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$
<br><br>Hence equation $$(iii)$$ is of order $$1$$ and degree $$3.$$ | mcq | aieee-2005 |
PVddRaYxCsHwkkYj | maths | differential-equations | order-and-degree | The differential equation whose solution is $$A{x^2} + B{y^2} = 1$$
<br/>where $$A$$ and $$B$$ are arbitrary constants is of | [{"identifier": "A", "content": "second order and second degree"}, {"identifier": "B", "content": "first order and second degree "}, {"identifier": "C", "content": "first order and first degree"}, {"identifier": "D", "content": "second order and first degree"}] | ["D"] | null | $$A{x^2} + B{y^2} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$Ax + by{{dy} \over {dx}} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>$$A + By{{{d^2}y} \over {d{x^2}}} + B{\left( {{{dy} \over {dx}}} \right)^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 3 \right)$$
<br><br>From $$(2)$$ and $$(3)$$
<br><br>$$x\left\{ { - By{{{d^2}y} \over {d{x^2}}} - B{{\left( {{{dy} \over {dx}}} \right)}^2}} \right\} + By{{dy} \over {dx}} = 0$$
<br><br>Dividing both sides by $$-B,$$ we get
<br><br>$$ \Rightarrow xy{{{d^2}y} \over {d{x^2}}} + x{\left( {{{dy} \over {dx}}} \right)^2} - y{{dy} \over {dx}} = 0$$
<br><br>Which is $$DE$$ of order $$2$$ and degree $$1.$$ | mcq | aieee-2006 |
4Alko7gVMaP2XraCHw1kluhs5rq | maths | differential-equations | order-and-degree | The difference between degree and order of a differential equation that represents the family of curves given by $${y^2} = a\left( {x + {{\sqrt a } \over 2}} \right)$$, a > 0 is _________. | [] | null | 2 | $${y^2} = a\left( {x + {{\sqrt a } \over 2}} \right)$$
<br><br>Differentiating both sides, we get
<br><br>$$2yy' = a$$<br><br>$${y^2} = 2yy'\left( {x + {{\sqrt {2yy'} } \over 2}} \right)$$<br><br>$$y = 2y'\left( {x + {{\sqrt {yy'} } \over {\sqrt 2 }}} \right)$$<br><br>$$y - 2xy' = \sqrt 2 y'\sqrt {yy'} $$<br><br>$${\left( {y - 2x{{dy} \over {dx}}} \right)^2} = 2y{\left( {{{dy} \over {dx}}} \right)^3}$$<br><br>D = 3 & O = 1<br><br>$$ \therefore $$ D $$-$$ O = 3 $$-$$ 1 = 2 | integer | jee-main-2021-online-26th-february-morning-slot |
lkfclsdOLEyMMUct | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The solution of the equation $$\,{{{d^2}y} \over {d{x^2}}} = {e^{ - 2x}}$$ | [{"identifier": "A", "content": "$${{{e^{ - 2x}}} \\over 4}$$ "}, {"identifier": "B", "content": "$${{{e^{ - 2x}}} \\over 4} + cx + d$$ "}, {"identifier": "C", "content": "$${1 \\over 4}{e^{ - 2x}} + c{x^2} + d$$ "}, {"identifier": "D", "content": "$$\\,{1 \\over 4}{e^{ - 4x}} + cx + d$$ "}] | ["B"] | null | $${{{a^2}y} \over {d{x^2}}} = {e^{ - 2x}};{\mkern 1mu} $$
<br><br>$${{dy} \over {dx}} = {{{e^{ - 2x}}} \over { - 2}} + c;$$
<br><br>$$y = {{{e^{ - 2x}}} \over 4} + cx + d$$ | mcq | aieee-2002 |
16rF0FrylxZOUJ23 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Solution of the differential equation $$ydx + \left( {x + {x^2}y} \right)dy = 0$$ is | [{"identifier": "A", "content": "$$log$$ $$y=Cx$$"}, {"identifier": "B", "content": "$$ - {1 \\over {xy}} + \\log y = C$$ "}, {"identifier": "C", "content": "$${1 \\over {xy}} + \\log y = C$$ "}, {"identifier": "D", "content": "$$ - {1 \\over {xy}} = C$$ "}] | ["B"] | null | $$ydx + \left( {x + {x^2}y} \right)dy = 0$$
<br><br>$$ \Rightarrow {{dx} \over {dy}} = - {x \over y} - {x^2}$$
<br><br>$$ \Rightarrow {{dx} \over {dy}} + {x \over y} = - {x^2},$$
<br><br>It is Bernoullis form. Divide by $${x^2}$$
<br><br>$${x^{ - 2}}{{dx} \over {dy}} + {x^{ - 1}}\left( {{1 \over y}} \right) = - 1.$$
<br><br>put $${x^{ - 1}} = t,\,\, - {x^{ - 2}}{{dx} \over {dy}} = {{dt} \over {dy}}$$
<br><br>We get, $$ - {{dt} \over {dy}} + t\left( {{1 \over y}} \right) = - 1$$
<br><br>$$ \Rightarrow {{dt} \over {dy}} - \left( {{1 \over y}} \right)t = 1$$
<br><br>It is linear in $$t.$$
<br><br>Integrating factor
<br><br>$$ = {e^{\int { - {1 \over y}dy} }} = {e^{ - \log y}} = {y^{ - 1}}$$
<br><br>$$\therefore$$ Solution is
<br><br>$$t\left( {{y^{ - 1}}} \right) = \int {\left( {{y^{ - 1}}} \right)} dy + c$$
<br><br>$$ \Rightarrow {1 \over x}.{1 \over y} = \log y + c$$
<br><br>$$ \Rightarrow \log y - {1 \over {xy}} = c$$ | mcq | aieee-2004 |
HRaARJOMPWqSETh4 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | If $$x{{dy} \over {dx}} = y\left( {\log y - \log x + 1} \right),$$ then the solution of the equation is : | [{"identifier": "A", "content": "$$y\\log \\left( {{x \\over y}} \\right) = cx$$ "}, {"identifier": "B", "content": "$$x\\log \\left( {{y \\over x}} \\right) = cy$$ "}, {"identifier": "C", "content": "$$\\log \\left( {{y \\over x}} \\right) = cx$$ "}, {"identifier": "D", "content": "$$\\log \\left( {{x \\over y}} \\right) = cy$$ "}] | ["C"] | null | $${{xdy} \over {dx}} = y\left( {\log y - \log x + 1} \right)$$
<br><br>$${{dy} \over {dx}} = {y \over x}\left( {\log \left( {{y \over x}} \right) + 1} \right)$$
<br><br>Put $$\,\,\,\,y = vx$$
<br><br>$${{dy} \over {dx}} = v + {{xdv} \over {dx}}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, \Rightarrow v + {{xdv} \over {dx}} = v\left( {\log v + 1} \right)$$
<br><br>$${{xdv} \over {dx}} = v\,\log \,v$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {{dv} \over {v\,\log \,v}} = {{dx} \over x}$$
<br><br>Put $$\,\,\,\,\log \,v = z$$
<br><br>$$ \Rightarrow {1 \over v}dv = dz$$
<br><br>$$ \Rightarrow {{dz} \over x} = {{dx} \over x}$$
<br><br>$$ \Rightarrow \ln \,z = \ln x + \ln \,c$$
<br><br>$$x = cx\,\,\,\,$$ or $$\,\,\,\,\,\log v = cx\,\,\,$$
<br><br>or $$\,\,\,\,$$ $$\log \left( {{y \over x}} \right) = cx.$$ | mcq | aieee-2005 |
D1AbY8Kr3F4ppiWw | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Solution of the differential equation
<br/><br>$$\cos x\,dy = y\left( {\sin x - y} \right)dx,\,\,0 < x <{\pi \over 2}$$ is :</br> | [{"identifier": "A", "content": "$$y\\sec x = \\tan x + c$$ "}, {"identifier": "B", "content": "$$y\\tan x = \\sec x + c$$ "}, {"identifier": "C", "content": "$$\\tan x = \\left( {\\sec x + c} \\right)y$$ "}, {"identifier": "D", "content": "$$\\sec x = \\left( {\\tan x + c} \\right)y$$ "}] | ["D"] | null | $$\cos xdy = y\left( {\sin x - y} \right)dx$$
<br><br>$${{dy} \over {dx}} = y\tan x - {y^2}\,\,\sec \,x$$
<br><br>$${1 \over {{y^2}}}{{dy} \over {dx}} - {1 \over y}\,\tan \,x = - \sec \,x\,....\left( i \right)$$
<br><br>Let $$\,\,\,\,{1 \over y} = t \Rightarrow - {1 \over {{y^2}}}{{dy} \over {dx}} = {{dt} \over {dx}}$$
<br><br>From equation $$(i)$$
<br><br>$$ - {{dt} \over {dx}} - \tan x = - \sec x$$
<br><br>$$ \Rightarrow {{dt} \over {dx}} + \left( {\tan x} \right)t = \sec x$$
<br><br>$${\rm I}.F. = {e^{\int {\tan xdx} }} = {\left( e \right)^{\log \left| {\sec x} \right|}}\sec x$$
<br><br>Solution $$:$$ $$\,\,t\left( {{\rm I}.F} \right) = \int {\left( {{\rm I}.F} \right)\sec xdx} $$
<br><br>$$ \Rightarrow {1 \over y}\sec x = \tan x + c$$ | mcq | aieee-2010 |
2QP8dFoEGMElvV1X | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | Let $$I$$ be the purchase value of an equipment and $$V(t)$$ be the value after it has been used for $$t$$ years. The value $$V(t)$$ depreciates at a rate given by differential equation $${{dv\left( t \right)} \over {dt}} = - k\left( {T - t} \right),$$ where $$k>0$$ is a constant and $$T$$ is the total life in years of the equipment. Then the scrap value $$V(T)$$ of the equipment is | [{"identifier": "A", "content": "$$I - {{k{T^2}} \\over 2}$$ "}, {"identifier": "B", "content": "$$I - {{k{{\\left( {T - t} \\right)}^2}} \\over 2}$$ "}, {"identifier": "C", "content": "$${e^{ - kT}}$$ "}, {"identifier": "D", "content": "$${T^2} - {1 \\over k}$$ "}] | ["A"] | null | $${{dV\left( t \right)} \over {dt}} = - k\left( {T - t} \right)$$
<br><br>$$ \Rightarrow \int {dVt} = - k\int {\left( {T - t} \right)} dt$$
<br><br>$$V\left( t \right) = {{k{{\left( {T - t} \right)}^2}} \over 2} + c$$
<br><br>$$V\left( 0 \right) = I \Rightarrow I = {{K{T^2}} \over 2} + C$$
<br><br>$$ \Rightarrow C = I - {{K{T^2}} \over 2}$$
<br><br>$$\therefore$$ $$\,\,\,\,\,V\left( T \right) = 0 + C = I - {{K{T^2}} \over 2}$$ | mcq | aieee-2011 |
8gKnz0SCzC0mtk67 | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | The population $$p$$ $$(t)$$ at time $$t$$ of a certain mouse species satisfies the differential equation $${{dp\left( t \right)} \over {dt}} = 0.5\,p\left( t \right) - 450.\,\,$$ If $$p(0)=850,$$ then the time at which the population becomes zero is : | [{"identifier": "A", "content": "$$2ln$$ $$18$$"}, {"identifier": "B", "content": "$$ln$$ $$9$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$$$ln$$ $$18$$"}, {"identifier": "D", "content": "$$ln$$ $$18$$"}] | ["A"] | null | Given differential equation is
<br><br>$${{dp\left( t \right)} \over {dt}} = 0.5p\left( t \right) - 450$$
<br><br>$$ \Rightarrow {{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 450$$
<br><br>$$ \Rightarrow {{dp\left( t \right)} \over {dt}} = {{p\left( t \right) - 900} \over 2}$$
<br><br>$$ \Rightarrow 2{{dp\left( t \right)} \over {dt}} = \left[ {900 - p\left( t \right)} \right]$$
<br><br>$$ \Rightarrow 2{{dp\left( t \right)} \over {900 - p\left( t \right)}} = - dt$$
<br><br>Integrate both sides, we get
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\int {{{dp\left( t \right)} \over {900 - p\left( t \right)}}} = \int {dt} $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Let\,\,900 - p\left( t \right) = u$$
<br><br>$$ \Rightarrow - dp\left( t \right) = du$$
<br><br>$$\therefore$$ $$\,\,\,\,\,\,$$ We have,
<br><br>$$2\int {{{du} \over u}} = \int {dt \Rightarrow 2\,\ln \,u = t + c} $$
<br><br>$$ \Rightarrow 2\ln \left[ {900 - p\left( t \right)} \right] = t + c$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ when $$t = 0,p\left( 0 \right) = 850$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$2\ln \left( {50} \right) = c$$
<br><br>$$ \Rightarrow 2\left[ {\ln \left( {{{900 - p\left( t \right)} \over {50}}} \right)} \right] = t$$
<br><br>$$ \Rightarrow 900 - p\left( t \right) = 50{e^{{t \over 2}}}$$
<br><br>$$ \Rightarrow p\left( t \right) = 900 - 50{e^{{t \over 2}}}$$
<br><br>Let $$p\left( {{t_1}} \right) = 0$$
<br><br>$$0 = 900 - 50{e^{{{{t_1}} \over 2}}}$$
<br><br>$$\therefore$$ $$\,\,\,\,\,\,{t_1} = 2\ln 18$$ | mcq | aieee-2012 |
DbbeUvxqz32HCdYr | maths | differential-equations | solution-of-differential-equations-by-method-of-separation-variables-and-homogeneous | At present, a firm is manufacturing $$2000$$ items. It is estimated that the rate of change of production P w.r.t. additional number of workers $$x$$ is given by $${{dp} \over {dx}} = 100 - 12\sqrt x .$$ If the firm employs $$25$$ more workers, then the new level of production of items is | [{"identifier": "A", "content": "$$2500$$"}, {"identifier": "B", "content": "$$3000$$ "}, {"identifier": "C", "content": "$$3500$$"}, {"identifier": "D", "content": "$$4500$$"}] | ["C"] | null | Given, Rate of change is $${{dp} \over {dx}} = 100 - 12\sqrt x $$
<br><br>$$ \Rightarrow dP = \left( {100 - 12\sqrt x } \right)dx$$
<br><br>By intergrating $$\int {dP = \int {\left( {100 - 12\sqrt x } \right)} } dx$$
<br><br>$$\int {dP} = \int {\left( {100 - 12\sqrt x } \right)} dx$$
<br><br>$$P = 100x - 8{x^{3/2}} + C$$
<br><br>Given, when $$x=0$$ then $$P=2000$$
<br><br>$$ \Rightarrow C = 2000$$
<br><br>Now when $$x$$$$=25$$
<br><br>then $$P = 100 \times 25 - 8 \times {\left( {25} \right)^{3/2}} + 2000$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,$$ $$=4500-1000$$
<br><br>$$ \Rightarrow P = 3500$$ | mcq | jee-main-2013-offline |
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