Question

Solve the following differential equation: $(x+y+1)\frac{dy}{dx}=1$.

Answer 1

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

$\frac{dy}{dx}=\frac{1}{x+y+1}$

Put, $u=x+y$ $⟹\frac{du}{dx}=1+\frac{dy}{dx}$

$=1+\frac{1}{u+1}=\frac{u+2}{u+1}$

$\left(\frac{u+1}{u+2}\right)du=dx$

$(1-\frac{1}{u+2})du=dx$

$\int (1-\frac{1}{u+2})du=\int dx$

$u-ln|u+2|+ln|C|=x$

$ln\left|\frac{C}{u+2}\right|=x-u$

$\frac{C}{u+2}=e^{x-u}$

$u+2=C{e}^{u-x}$

$⟹(x+y)+2=C{e}^{(x+y)-x}$

$x=C{e}^y-y-2$