Question

Solve:-
$(e^x+1)ydy+(y+1)dx=0$

Answer 1

$(e^x+1)ydy+(y+1)dx=0$

$(e^x+1)ydy=-(y+1)dx$

$\frac{ydy}{y+1}=-\frac{dx}{e^x+1}$

$\frac{(y+1-1)dy}{y+1}=-\frac{dx}{e^x(1+e^{-x})}$

$\frac{(y+1)dy}{y+1}-\frac{dy}{y+1}=-\frac{e^{-x}dx}{(1+e^{-x})}$

$dy-\frac{dy}{y+1}=-\frac{e^{-x}dx}{(1+e^{-x})}$

Integrating both sides, we get

$\int dy-\int \frac{dy}{y+1}=\int \frac{-e^{-x}dx}{(1+e^{-x})}$

Let $t=1+e^{-x}\Rightarrow dt=-e^{-x}dx$

$y-\log |y+1|=\int \frac{dt}{t}$

$y-\log |y+1|=\log \left|t\right|$

$y-\log |y+1|=\log |1+e^{-x}|+c$ where $t=1+e^{-x}$

$y-\log |y+1|-\log |1+e^{-x}|-c=0$ is the required solution.