JEE Main 2025 Apr 8 Shift 2, Mathematics Q17: Methods of Solving First Order, First Degree Differential Equation

JEE Main2025Apr 8, Shift 2Mathematics

Let $f (x) = x - 1$ and $g (x) = e^{x}$ for $x \in R$ . If $\frac{d y}{d x} = (e^{- 2 x} g (f (f (x))) - \frac{y}{x})$ , $y (0) = 0$ , then $y (1)$ is:

A
$\frac{1 - e ^{2}}{e ^{4}}$
B
$\frac{2 e - 1}{e ^{3}}$
C
$\frac{e - 1}{e ^{4}}$
D
$\frac{1 - e ^{3}}{e ^{4}}$

Solution

Compute the composition: $f (f (x)) = f (x - 1) = x - 2$ , so $g (f (f (x))) = e^{x - 2}$ .

The ODE becomes $\frac{d y}{d x} + \frac{1}{x} y = e^{- 2 x} \cdot e^{x - 2} = e^{x - 2 - 2 x}$ , which is linear first-order.

Integrating factor: $μ = exp (\int \frac{d x}{x}) = e^{2 x}$ .

Multiplying through: $\frac{d}{d x} (y e^{2 x}) = e^{2 x} \cdot e^{x - 2 - 2 x} = e^{x - 2}$ .

Integrating: $y e^{2 x} = e^{x - 2} + C$ .

Apply $y (0) = 0$ : $0 = e^{- 2} + C$ , so $C = - e^{- 2}$ ; then $y e^{2 x} = e^{x - 2} - e^{- 2}$ .

At $x = 1$ : $y \cdot e^{2} = e^{- 1} - e^{- 2}$ , so $y = \frac{e ^{- 1} - e ^{- 2}}{e ^{2}} = \frac{1/ e - 1/ e ^{2}}{e ^{2}} = \frac{e - 1}{e ^{4}}$ , matching option (3).

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JEE Main 2025 Apr 8 Shift 2, Mathematics Q17: Methods of Solving First Order, First Degree Differential Equation

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