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Differential Equations question

2021 · 25 Jul · Shift 2 · Q40

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JEE Main
Mathematics
Differential Equations
2021 · 25 Jul · Shift 2 · Q40

Differential Equations question

2021 · 25 Jul · Shift 2 · Q40

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

Let y = y(x) be the solution of the differential equation xdy = (y + x3 cosx)dx with y(

\pi

) = 0, then

y\left( {{\pi \over 2}} \right)

is equal to :

A
${{{\pi ^2}} \over 4} + {\pi \over 2}$
B
${{{\pi ^2}} \over 2} + {\pi \over 4}$
C
${{{\pi ^2}} \over 2} - {\pi \over 4}$
D
${{{\pi ^4}} \over 4} - {\pi \over 2}$

View written solutionFree

Correct answer: A

Rewrite the differential equation

Given $x\,dy=(y+x^3\cos x)\,dx$

Divide by $dx$ : $x\frac{dy}{dx}=y+x^3\cos x$

So, $\frac{dy}{dx}-\frac{1}{x}y=x^2\cos x$

This is a linear differential equation: $\frac{dy}{dx}+P(x)y=Q(x)$ with $P(x)=-\frac{1}{x},\qquad Q(x)=x^2\cos x.$

Find the integrating factor

$\text{I.F.}=e^{\int -\frac{1}{x}\,dx}=e^{-\ln x}=\frac{1}{x}$

Multiply the equation by the integrating factor

Multiplying by $\frac{1}{x}$ : $\frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=x\cos x$

The left side becomes: $\frac{d}{dx}\left(\frac{y}{x}\right)=x\cos x$

So, $\frac{y}{x}=\int x\cos x\,dx+C$

Evaluate the integral

Using integration by parts: $\int x\cos x\,dx=x\sin x-\int \sin x\,dx=x\sin x+\cos x$

Hence, $\frac{y}{x}=x\sin x+\cos x+C$

Therefore, $y=x(x\sin x+\cos x+C)$

Use the initial condition $y(\pi)=0$

Substitute $x=\pi$ , $y=0$ : $0=\pi(\pi\sin \pi+\cos \pi+C)$

Since $\sin \pi=0,\qquad \cos \pi=-1$ we get $0=\pi(-1+C)$

Thus, $C=1$

So the solution is $y=x(x\sin x+\cos x+1)$

Find $y\left(\frac{\pi}{2}\right)$

$y\left(\frac{\pi}{2}\right)=\frac{\pi}{2}\left(\frac{\pi}{2}\sin\frac{\pi}{2}+\cos\frac{\pi}{2}+1\right)$

Now, $\sin\frac{\pi}{2}=1,\qquad \cos\frac{\pi}{2}=0$

So, $y\left(\frac{\pi}{2}\right)=\frac{\pi}{2}\left(\frac{\pi}{2}+1\right)$

$=\frac{\pi^2}{4}+\frac{\pi}{2}$

Compare with options

This matches: $\boxed{\frac{\pi^2}{4}+\frac{\pi}{2}}$ which is Option A.

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