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

<p>$$\begin{aligned} & f(x)=x-1 \\ & f(f(x))=f(x)-1=x-1-1=x-2 \\ & g(f(f(x)))=e^{x-2} \\ & \therefore \frac{d y}{d x}=e^{-2 \sqrt{x}} \times e^{x-2}-\frac{1}{\sqrt{x}} y \\ & \frac{d y}{d x}+\frac{1}{\sqrt{x}} y=e^{x-2 \sqrt{x}-2} \text { which is L.D.E } \\ & \text { I.F. }=e^{\int \frac{d y}{\sqrt{x}}}=e^{2 \sqrt{x}} \end{aligned}$$</p> <p>Its solution is</p> <p>$$\begin{aligned} & y \times e^{2 \sqrt{x}}=\int e^{2 \sqrt{x}} \times e^{x-2 \sqrt{x}-2} d x+c \\ & y \times e^{2 \sqrt{x}}=\int e^{x-2} d x+c \\ & y \times e^{2 \sqrt{x}}=e^{x-2}+c \end{aligned}$$</p> <p>Given $\mathrm{x}=0, \mathrm{y}=0 \Rightarrow 0=\mathrm{e}^{-2}+\mathrm{c} \quad ; \mathrm{c}=-\mathrm{e}^{-2}$</p> <p>$$\therefore \mathrm{y} \times \mathrm{e}^{2 \sqrt{x}}=\mathrm{e}^{\mathrm{x}-2}-\mathrm{e}^{-2}$$</p> <p>when $\mathrm{x}=1, \mathrm{y} \times \mathrm{e}^2=\mathrm{e}^{-1}-\mathrm{e}^{-2}$</p> <p>$$y=\frac{e^{-1}-e^{-2}}{e^2}=\frac{\frac{1}{e}-\frac{1}{e^2}}{e^2}=\frac{e^2-e}{e^5}=\frac{e-1}{e^4}$$</p> <p>Option (1) is correct</p>