Question

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

Answer 1

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

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

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

let $1+y^2=t$

then $2ydy=dt$

$\therefore \int \left(\frac{1}{t}\right)\left(\frac{dt}{2}\right)=x+c$

$\left(\frac{1}{2}\right)\log t=x+c$

$\log (1+y^2)=2x+c_1$