Let g be a differentiable function such that $\int_{0}^{\mathrm{x}} \mathrm{g}(\mathrm{t}) \mathrm{dt}=\mathrm{x}-\int_{0}^{\mathrm{x}} \mathrm{tg}(\mathrm{t}) \mathrm{dt}$, $\mathrm{x} \geq 0$ and let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ satisfy the differential equation $\frac{dy}{dx}-y\tan x= 2(x+1)\sec x\, g(x)$, $x \in\left[0, \frac{\pi}{2}\right)$. If $y(0)=0$, then $\mathrm{y}\left(\frac{\pi}{3}\right)$ is equal to

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Differential Equations

Let g be a differentiable function such that

\int_{0}^{x} g (t) dt = x - \int_{0}^{x} tg (t) dt

x \geq 0

and let

y = y (x)

satisfy the differential equation

\frac{d y}{d x} - y tan x = 2 (x + 1) sec x g (x)

x \in [0, \frac{π}{2})

. If

y (0) = 0

, then

y (\frac{π}{3})

is equal to

Answer: (B)

Differentiating

\int_{0}^{x} g (t) d t = x - \int_{0}^{x} t g (t) d t

gives

g (x) = 1 - x g (x)

, so

g (x) = \frac{1}{1 + x}

. The equation becomes

\frac{d y}{d x} - y tan x = 2 (x + 1) sec x \cdot \frac{1}{1 + x} = 2 sec x

. The integrating factor is

e^{- \int t a n x d x} = cos x

, so

\frac{d}{d x} (y cos x) = 2 sec x cos x = 2

. Integrating,

y cos x = 2 x + c

, and

y (0) = 0

forces

c = 0

, giving

y = 2 x sec x

. At

x = \frac{π}{3}

y = 2 \cdot \frac{π}{3} \cdot 2 = \frac{4 π}{3}

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option