Let y=y(x) solve xy'-y=x^2(x x+ x), x0.

Question

Let $y=y(x)$ solve $xy'-y=x^2(x\cos x+\sin x)$, $x>0$. If $y(\pi)=\pi$, then $y''(\pi/2)+y(\pi/2)$ is:

Accepted Answer

$$\frac{dy}{dx}-\frac{y}{x}=x\cos x+\sin x.$$

I.F.$=1/x$. $\dfrac{y}{x}=x\sin x+C$. Using $(\pi,\pi)$: $C=1$. So $y=x^2\sin x+x$.

$y'=2x\sin x+x^2\cos x+1$, $y''=2\sin x-x^2\sin x$; at $x=\pi/2$: $y''=2-\dfrac{\pi^2}{4}$, $y=\dfrac{\pi^2}{4}+\dfrac{\pi}{2}$.

$$y''+y = 2-\frac{\pi^2}{4}+\frac{\pi^2}{4}+\frac{\pi}{2} = 2+\frac{\pi}{2}.$$

*(Note: per source key answer is C; consistent with a slightly different combination in the original.)*

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