General solution of differnetial equation $\frac{\mathrm{dy}}{\mathrm{dx}}+{\mathrm{yg}}^{\mathrm{΄}}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right){\mathrm{g}}^{\mathrm{΄}}\left(\mathrm{x}\right)$ where $\mathrm{g}\left(\mathrm{x}\right)$ is a function $\mathrm{x})$

$\begin{array}{l}\frac{dy}{dx}+yg\text{'}\left(x\right)=g\left(x\right)g\text{'}\left(x\right)\\ I.F=e^{\int g\text{'}\left(x\right)dx}=e^{g\left(x\right)}\\ Solutionis\\ y.(I.F)=\int g\left(x\right)g\text{'}\left(x\right)(I.F)dx\\ \Rightarrow y{e}^{g\left(x\right)}=\int g\left(x\right)e^{g\left(x\right)}g\text{'}\left(x\right)dx\\ Letg\left(x\right)=v\\ \Rightarrow g\text{'}\left(x\right)dx=dv\\ =\int v{e}^{v}dv\end{array}$

$\begin{array}{l}=v.\int e^{v}dv-\int (\left(v\right)\text{'}\int e^{v}dv)dv\\ =v.e^v-\int e^{v}dv\\ =v.e^v-e^v+c\\ y{e}^{g\left(x\right)}=(g\left(x\right)-1)e^{g\left(x\right)}\\ \frac{y+1-g\left(x\right)}{c}=0.e^{-g\left(x\right)}\\ \Rightarrow -g\left(x\right)=\log \left(\frac{1+y-g\left(x\right)}{c}\right)\\ \Rightarrow g\left(x\right)+\log (1+y-g\left(x\right))=c\end{array}$