Question

Let $y=y\left(x\right)$ be solution of the differential equation,
$x{y}^{\prime }-y=x^2(x\cos x+\sin x),x>0.$ If $y\left(\pi \right)=\pi ,$ then $y^{\prime \prime }\left(\frac{\pi }{2}\right)+y\left(\frac{\pi }{2}\right)$ is equal to

• A. $2+\frac{\pi }{2}+\frac{{\pi }^2}{4}$
• B. $2+\frac{\pi }{2}$
• C. $1+\frac{\pi }{2}$
• D. $1+\frac{\pi }{2}+\frac{{\pi }^2}{4}$

Answer 1

The correct option is B $2+\frac{\pi }{2}$

Given: $x{y}^{\prime }-y=x^2(x\cos x+\sin x)$
$\Rightarrow y^{\prime }-\frac{y}{x}=x(x\cos x+\sin x)$
Which is linear equation.
$\therefore$ $I.F.=$$e^{-\int \left(1/x\right)dx}=e^{-\ln x}=\frac{1}{x}$
Now, the general solution,
$y\cdot \frac{1}{x}=\int \frac{1}{x}.x(x\cos x+\sin x)dx$
$\Rightarrow \frac{y}{x}=\int \frac{d}{dx}\left(x\sin x\right)dx$
$\Rightarrow \frac{y}{x}=x\sin x+C$
$\Rightarrow y=x^2\sin x+Cx$
Since $y\left(\pi \right)=\pi \Rightarrow C=1$
$\therefore y=x^2\sin x+x\Rightarrow y\left(\frac{\pi }{2}\right)=\frac{{\pi }^2}{4}+\frac{\pi }{2}$

$\Rightarrow y^{\prime }=2x\sin x+x^2\cos x+1$
$\Rightarrow y^{\prime \prime }=2\sin x+2x\cos x+2x\cos x-x^2\sin x$
$\Rightarrow y^{\prime \prime }=2\sin x+4x\cos x-x^2\sin x$
$\Rightarrow y^{\prime \prime }\left(\frac{\pi }{2}\right)=2-\frac{{\pi }^2}{4}$
$\therefore y\left(\frac{\pi }{2}\right)+y^{\prime \prime }\left(\frac{\pi }{2}\right)=2+\frac{\pi }{2}$