JEE Main 2025 Apr 3 Shift 1, Mathematics Q12: Methods of Solving First Order, First Degree Differential Equation

JEE Main2025Apr 3, Shift 1Mathematics

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

A
$\frac{2 π}{3 3}$
B
$\frac{4 π}{3}$
C
$\frac{2 π}{3}$
D
$\frac{4 π}{3 3}$

Solution

Differentiating $\int_{0}^{x} g (t) d t = x - \int_{0}^{x} t g (t) d t$ with respect to $x$ gives $g (x) = 1 - x g (x)$ , so

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

The ODE becomes $y^{'} - y tan x = 2 (x + 1) sec x \cdot \frac{1}{x + 1} = 2 sec x .$

Integrating factor: $exp (- \int tan x d x) = exp (ln cos x) = cos x .$

Multiplying through: $(y cos x)^{'} = 2 sec x \cdot cos x = 2.$

Integrate: $y cos x = 2 x + C$ . With $y (0) = 0$ , $C = 0$ , so $y = \frac{2 x}{cos x} = 2 x sec x .$

At $x = \frac{π}{3}$ : $sec \frac{π}{3} = 2$ , hence $y (\frac{π}{3}) = 2 \cdot \frac{π}{3} \cdot 2 = \frac{4 π}{3} .$

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JEE Main 2025 Apr 3 Shift 1, Mathematics Q12: Methods of Solving First Order, First Degree Differential Equation

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