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

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Differential Equations

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 :-

Answer: (C)

Here

f (f (x)) = f (x - 1) = x - 2

, so

g (f (f (x))) = e^{x - 2}

and the equation becomes

\frac{d y}{d x} + \frac{1}{x} y = e^{x - 2 x - 2}

. The integrating factor is

e^{\int x^{- 1/2} d x} = e^{2 x}

, so

\frac{d}{d x} (y e^{2 x}) = e^{2 x} \cdot e^{x - 2 x - 2} = e^{x - 2}

. Integrating gives

y e^{2 x} = e^{x - 2} + C

, and

y (0) = 0

forces

C = - e^{- 2}

. At

x = 1

y e^{2} = e^{- 1} - e^{- 2} = \frac{e - 1}{e ^{2}}

, so

y (1) = \frac{e - 1}{e ^{4}}

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option

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