JEE PYQ: Differential Equation Question 21
Question 21 - 2020 (04 Sep Shift 1)
Let $y = y(x)$ be the solution of the differential equation $xy’ - y = x^2(x\cos x + \sin x)$, $x > 0$. If $y(\pi) = \pi$, then $y’’\left(\frac{\pi}{2}\right) + y\left(\frac{\pi}{2}\right)$ is equal to:
(1) $2 + \frac{\pi}{2}$
(2) $1 + \frac{\pi}{2} + \frac{\pi^2}{4}$
(3) $2 + \frac{\pi}{2} + \frac{\pi^2}{4}$
(4) $1 + \frac{\pi}{2}$
Show Answer
Answer: (1)
Solution
$\frac{dy}{dx} - \frac{y}{x} = x(x\cos x + \sin x)$. I.F. $= \frac{1}{x}$. So $\frac{y}{x} = \int (x\cos x + \sin x),dx = x\sin x + C$. $y(\pi) = \pi$: $1 = 0 + C$, $C = 1$. So $\frac{y}{x} = x\sin x + 1$, $y = x^2\sin x + x$. $y\left(\frac{\pi}{2}\right) = \frac{\pi^2}{4} + \frac{\pi}{2}$. $y’ = 2x\sin x + x^2\cos x + 1$. $y’’ = 2\sin x + 4x\cos x - x^2\sin x$. $y’’\left(\frac{\pi}{2}\right) = 2 + 0 - \frac{\pi^2}{4}$. Sum $= 2 - \frac{\pi^2}{4} + \frac{\pi^2}{4} + \frac{\pi}{2} = 2 + \frac{\pi}{2}$.
BETA
