Question

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

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

Hint:

Divide by \(x^2\): \(\frac{xy^{\prime}-y}{x^2}=\frac{d}{dx}\left(\frac{y}{x}\right)\). Integrate, use \(y(\pi)=\pi\), then compute \(y^{\prime\prime}\) and \(y\) at \(\pi/2\).

Answer 1

The Correct Option is C

Dividing by \(x^2\): \(\frac{d}{dx}\left(\frac{y}{x}\right)=x\cos x+\sin x=\frac{d}{dx}(x\sin x)\).
So \(\frac{y}{x}=x\sin x+C\Rightarrow y=x^2\sin x+Cx\). \(y(\pi)=\pi\Rightarrow C\pi=\pi\Rightarrow C=1\).
\(y=x^2\sin x+x\). \(y^{\prime}=2x\sin x+x^2\cos x+1\). \(y^{\prime\prime}=2\sin x+4x\cos x-x^2\sin x\).
At \(x=\frac{\pi}{2}\): \(y^{\prime\prime}=2+0-\frac{\pi^2}{4}\), \(y=\frac{\pi^2}{4}+\frac{\pi}{2}\).
Sum \(=2-\frac{\pi^2}{4}+\frac{\pi^2}{4}+\frac{\pi}{2}=2+\frac{\pi}{2}\).
Answer: (3) \(2+\frac{\pi}{2}\)