Question

$y\frac{dy}{dx}=1+y^2$

Answer 1

Consider the given deferential equation,

$y\frac{dy}{dx}=1+y^2$

$\frac{y}{1+y^2}dy=dx$

$\frac{2y}{2(1+y^2)}dy=dx$

Taking integration both sides,

$\int \frac{2y}{2(1+y^2)}dy=\int 1dx$

$\frac{1}{2}\int \frac{2y}{(1+y^2)}dy=\int 1dx$

$\frac{1}{2}\log (1+y^2)=x+C$

$\log {(1+y^2)}^{\frac{1}{2}}=x+C$

$\log \sqrt{(1+y^2)}=x+C$

Hence, this is the answer.