Let $y=y(x)$ satisfies the equation $\frac{d y}{d x}-|A|=0$, for all $x>0$, where $A=\left[\begin{array}{ccc}y & \sin x & 1 \\ 0 & -1 & 1 \\ 2 & 0 & \frac{1}{x}\end{array}\right]$. If $y(\pi)=\pi+2$, then the value of $y\left(\frac{\pi}{2}\right)$ is :

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

(B) $\frac{3 \pi}{2}-\frac{1}{\pi}$

(C) $\frac{\pi}{2}-\frac{4}{\pi}$

(D) $\frac{\pi}{2}-\frac{1}{\pi}$

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Differential Equations

Let

y = y (x)

satisfies the equation

\frac{d y}{d x} - ∣ A ∣ = 0

, for all

x > 0

, where

A = y 02 sin x - 1 0 11 \frac{1}{x}

. If

y (π) = π + 2

, then the value of

y (\frac{π}{2})

is :

Answer: (A)

Expanding the determinant along the first row,

∣ A ∣ = y (- \frac{1}{x} - 0) - sin x (0 - 2) + 1 (0 + 2) = - \frac{y}{x} + 2 sin x + 2

so the differential equation is

\frac{d y}{d x} + \frac{y}{x} = 2 sin x + 2

. Multiplying by

x

makes the left side an exact derivative:

\frac{d}{d x} (x y) = 2 x sin x + 2 x ⟹ x y = - 2 x cos x + 2 sin x + x^{2} + c

x = π

with

y = π + 2

this gives

π^{2} + 2 π = 2 π + 0 + π^{2} + c

, so

c = 0

. Putting

x = \frac{π}{2}

\frac{π}{2} y = 0 + 2 + \frac{π ^{2}}{4} ⟹ y = \frac{4}{π} + \frac{π}{2}

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

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option