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

(A) $1+\frac{\pi}{2}+\frac{\pi^{2}}{4}$

(B) $2+\frac{\pi}{2}+\frac{\pi^{2}}{4}$

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation,

x y^{'} - y = x^{2} (x cos x + sin x), x > 0

. If

y (π) = π

, then

y^{''} (\frac{π}{2}) + y (\frac{π}{2})

is equal to:

Answer: (C)

Write the equation as

y^{'} - \frac{y}{x} = x (x cos x + sin x)

. The integrating factor is

e^{- \int d x / x} = \frac{1}{x}

, and multiplying through turns the left side into

\frac{d}{d x} (\frac{y}{x}) = x cos x + sin x

. Integrating,

\frac{y}{x} = x sin x + C

. The condition

y (π) = π

gives

1 = π sin π + C

, so

C = 1

and

y = x^{2} sin x + x

. Differentiating twice,

y^{''} = - x^{2} sin x + 4 x cos x + 2 sin x

, so at

x = \frac{π}{2}

y (\frac{π}{2}) = \frac{π ^{2}}{4} + \frac{π}{2}

and

y^{''} (\frac{π}{2}) = - \frac{π ^{2}}{4} + 2

. Adding,

y^{''} (\frac{π}{2}) + y (\frac{π}{2}) = 2 + \frac{π}{2}

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option