Question

# Solve differential equationy'+y/x= 3y^(-2) x>0

First order differential equations

Solve differential equation $$y'+y/x= 3y^{-2}$$ $$x>0$$

2020-12-14

Dividing both sidesby $$y^{-2}$$
$$y^2y'+ y^3/x=3$$
$$u=y^3$$
Differentiating both sides with respect to x
$$du/dx= 3y^2 dy/dx$$
$$1/3 (du)/dx= y^2 dy/dx$$
Substituting this in the equation $$y^2y'+ y^3/x=3$$
Substituting $$1/3 (du)/dx= y^2 dy/dx$$ and $$u=y^3$$ in the equation $$y^2y'+ y^3/x=3$$
$$1/3 (du)/dx+ u/x=3$$
$$1/9 (du)/dx+u/(3x)=1$$