Let f(x)=x-1 and g(x)=e x for x in mathbb R . If d y/d x= (e -2 √(x) g(f(f(x)))-frac y √(x) ), y(0)=0 , then y(1) is

Mathematics · Differential Equations

JEE Main 2025 — 8 April, Evening Shift — Question 34

Let $f(x)=x-1$ and $g(x)=e^{x}$ for $x \in \mathbb{R}$ . 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

$f(x)=x-1$

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

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

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

$\Rightarrow \frac{d y}{d x}+\frac{y}{\sqrt{x}}=e^{-2 \sqrt{x}} e^{x-2}$

IF $=e^{\int x^{-12} d x}=e^{2 \sqrt{x}}$

$y . e^{2 \sqrt{x}}=\int e^{-2 \sqrt{x}} e^{x-2} e^{2 \sqrt{x}} d x$

$=y \cdot e^{2 \sqrt{x}}=e^{x-2}+c$

$y(0)=0 \Rightarrow c=e^{-2}$

$\therefore y \cdot e^{2 \sqrt{x}}=e^{x-2}-e^{-2}$

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

$y=\frac{e-1}{e^{4}}$

Answer key and solution verified before publishing.

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