If $\frac{d y}{d x}+2 y \tan x=\sin x, 0<x<\frac{\pi}{2}$ and $y\left(\frac{\pi}{3}\right)=0$, then the maximum value of $\mathrm{y}(\mathrm{x})$ is

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Differential Equations

\frac{d y}{d x} + 2 y tan x = sin x, 0 < x < \frac{π}{2}

and

y (\frac{π}{3}) = 0

, then the maximum value of

y (x)

Answer: (A)

This is a first-order linear ODE. The integrating factor is:

I . F . = e^{\int 2 t a n x d x} = e^{2 l n ∣ s e c x ∣} = sec^{2} x

Multiplying through:

y sec^{2} x = \int sin x sec^{2} x d x = \int tan x sec x d x = sec x + C

.

Using

y (π /3) = 0

0 = sec (π /3) + C = 2 + C

, so

C = - 2

.

Thus

y = cos x - 2 cos^{2} x = - 2 (cos x - \frac{1}{4})^{2} + \frac{1}{8}

.

The maximum value is

\frac{1}{8}

when

cos x = \frac{1}{4}

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option

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