Question

Solve the differential equations dy/dx=((xy^2)+x)/y

Answer 1

$\displaystyle \dfrac{dy}{dx}=\dfrac{xy^2+x}{y}\\ \dfrac{dy}{dx}=\dfrac{x(y^2+1)}{y}\\ \dfrac{y}{y^2+1}\, dy=x\, dx\\ \int \dfrac{y}{y^2+1}\, dy=\int x\, dx\\ \dfrac{\ln (y^2+1)}{2}=\dfrac{x^2}{2}+C\\ \ln(y^2+1)=x^2+C\\ y^2+1=e^{x^2+C}\\ y^2=e^{x^2+C}-1\\ y=\sqrt{e^{x^2+C}-1} \vee y=-\sqrt{e^{x^2+C}-1}\\ \boxed{y=\sqrt{Ce^{x^2}-1} \vee y=-\sqrt{Ce^{x^2}-1}}$