Thank you for reporting, we will resolve it shortly
Q.
General solution of the differential equation $\frac{ dy }{ dx }+ yg'( x )= g ( x ) \cdot g '( x )$, where $g ( x )$ is a function of $x$ is
Differential Equations
Solution:
We have, $\frac{d y}{d x}=[g(x)-y] g'(x)$
Put $g(x)-y=V $
$\Rightarrow g'(x)-\frac{d y}{d x}=\frac{d V}{d x}$
Hence, $g'(x)-\frac{d V}{d x}=V \cdot g'(x)$
$\Rightarrow \frac{ d V }{ dx }=(1- V ) \cdot g'( x ) $
$\Rightarrow \frac{ d V }{1- V }= g'( x ) d x$
$\Rightarrow \int \frac{ dV }{1- V }=\int g '( x ) dx $
$\Rightarrow -\log (1- V )= g ( x )- C$
$\Rightarrow g ( x )+\log (1- V )= C$
$\therefore g(x)+\log [1+y-g(x)]=C$