General solution of the differential equation  $\frac{dy}{dx}+y{g}^{\mathrm{\prime }}\left(x\right)=g\left(x\right)\cdot g^{\mathrm{\prime }}\left(x\right)$, where g(x) is a function of x is

We have, $\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\mathrm{g}\right(\mathrm{x})-\mathrm{y}]{\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)$

Put $g\left(x\right)-y=V\Rightarrow g^{\mathrm{\prime }}\left(x\right)-\frac{dy}{dx}=\frac{dV}{dx}$

Hence, $g^{\mathrm{\prime }}\left(x\right)-\frac{dV}{dx}=V\cdot g^{\mathrm{\prime }}\left(x\right)$

$\begin{array}{l}\Rightarrow \frac{\mathrm{dV}}{\mathrm{dx}}=(1-\mathrm{V})\cdot {\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\Rightarrow \frac{\mathrm{dV}}{1-\mathrm{V}}={\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\mathrm{dx}\\ \Rightarrow \int \frac{\mathrm{dV}}{1-\mathrm{V}}=\int {\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\mathrm{dx}\Rightarrow -\log (1-\mathrm{V})=\mathrm{g}\left(\mathrm{x}\right)-\mathrm{C}\\ \Rightarrow \mathrm{g}\left(\mathrm{x}\right)+\log (1-\mathrm{V})=\mathrm{C}\\ \therefore \mathrm{g}\left(\mathrm{x}\right)+\log [1+\mathrm{y}-\mathrm{g}(\mathrm{x}\left)\right]=\mathrm{C}\end{array}$