Question

Solve the differential equation $(e^x+1)ydy+(y+1)dx=0$

Answer 1

We have,

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

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

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

On taking integration both sides, we get

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

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

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

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

Let $t=e^x$

$dt=e^{x}dx$

Therefore,

$I=-\int \frac{1}{t(t+1)}dt$

$I=-\int (\frac{1}{t}-\frac{1}{t+1})dt$

$I=-(\ln \left(t\right)-\ln (t+1))$

$I=\ln (t+1)-\ln \left(t\right)$

$I=\ln (e^x+1)-\ln \left(e^x\right)$

$I=\ln \left(\frac{e^x+1}{e^x}\right)$

From equation (1),

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

$y-\ln (y+1)=\ln \left(\frac{e^x+1}{e^x}\right)+C$

Hence, this is the answer.