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

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