The general solution of the differential equation $(e^x+1)ydy=(y+1)e^xdx$ is

$(A)\;(y+1)=k(e^x+1) \quad (B)\;y+1=e^x+1+k \quad (C)\;y=log\{k(y+1)(e^x+1)\} \quad (D)\;ylog{\frac{e^x+1}{y+1}}+k$

Toolbox:
• A linear differential equation is of the form $\large\frac{dy}{dx}$$=f(x), then it can be solved by seperating the variables. Step 1: Given (e^x+1)y\;dy=(y+1)e^xdx Seperating the variables we get, \large\frac{y\;dy}{y+1}=\large\frac{e^x}{e^x+1}$$dx$
On integrtaing we get,
$\int \large\frac{y\;dy}{y+1}=\int \large\frac{e^x}{e^x+1}$$dx Consider the L.H.S Add and subtract 1 to the numerator \int \large\frac{(y+1)-1}{y+1}$$dy$
On seperating the terms we get,
$\int dy-\int \large\frac{1}{y+1}$$dy => y-\log(y+1) Step 2: Now Consider R.H.S \int \large \frac{e^x}{e^x+1}$$dx$
Put $e^x+1=t$ On differentiating w.r.t x
$e^xdx=dt$ on substituting this