Solve the differential equation

$\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1$

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

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

This equation is a linear differential equation of the form

$\frac{d y}{d x}+P y=Q$, where $P=\frac{1}{\sqrt{x}}$ and $Q=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}} .$

Now, I.F $=e^{\int P d x}=e^{\int \frac{1}{\sqrt{x}} d x}=e^{2 \sqrt{x}}$

$\frac{d y}{d x}+P y=Q$, where $P=\frac{1}{\sqrt{x}}$ and $Q=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}} .$

The general solution of the given differential equation is given by,

$y($ I.F. $)=\int($ Q $\times$ I.F. $) d x+$ C

$\Rightarrow y e^{2 \sqrt{x}}=\int\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}} \times e^{2 \sqrt{x}}\right) d x+$ C

$\Rightarrow y e^{2 \sqrt{x}}=\int \frac{1}{\sqrt{x}} d x+$ C

$\Rightarrow y e^{2 \sqrt{x}}=2 \sqrt{x}+$ C