Let $y=y(x)$ be the solution of the differential equation $2 \cos \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=\sin 2 \mathrm{x}-4 \mathrm{y} \sin \mathrm{x}, \mathrm{x} \in\left(0, \frac{\pi}{2}\right)$. If $y\left(\frac{\pi}{3}\right)=0$, then $y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)$ is equal to ____.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

2 cos x \frac{dy}{dx} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2})

. If

y (\frac{π}{3}) = 0

, then

y^{'} (\frac{π}{4}) + y (\frac{π}{4})

is equal to ____.

Answer: 1

Starting from

2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x

, divide through by

2 cos x

and use

sin 2 x = 2 sin x cos x

\frac{d y}{d x} = sin x - 2 y tan x ⟹ \frac{d y}{d x} + 2 y tan x = sin x .

This is linear in

y

. 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 by

sec^{2} x

\frac{d}{d x} (y sec^{2} x) = sin x sec^{2} x = tan x sec x .

Integrating both sides:

y sec^{2} x = sec x + C ⟹ y = cos x + C cos^{2} x .

Using

y (\frac{π}{3}) = 0

cos \frac{π}{3} = \frac{1}{2}

cos^{2} \frac{π}{3} = \frac{1}{4}

, so

0 = \frac{1}{2} + \frac{C}{4} \Rightarrow C = - 2.

y = cos x - 2 cos^{2} x

, and differentiating,

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

x = \frac{π}{4}

cos \frac{π}{4} = sin \frac{π}{4} = \frac{1}{2}

, so

y (\frac{π}{4}) = \frac{1}{2} - 2 \cdot \frac{1}{2} = \frac{1}{2} - 1,

y^{'} (\frac{π}{4}) = - \frac{1}{2} + 2 sin \frac{π}{2} = - \frac{1}{2} + 2.

Adding:

y^{'} (\frac{π}{4}) + y (\frac{π}{4}) = (- \frac{1}{2} + 2) + (\frac{1}{2} - 1) = 1.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Integer answer