Solve the differential equation $\left[ {\frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }} - \frac{y}{{\sqrt x }}} \right]\frac{{dx}}{{dy}} = 1(x \ne 0)$

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

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

This equation is a linear differential equation of the form $\frac{{dy}}{{dx}} + Py = Q$, where $P = \frac{1}{{\sqrt x }}$ and $Q = \frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }}$.

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

The general solution of the given differential equation is given by, $y({\rm{I}}.{\rm{F}}.) = \int {({\rm{Q}} \times {\rm{I}}.{\rm{F}}.)} dx + {\rm{C}}$

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

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