Question

Solve the differential equation.

x y^2 y' = x + 1

Answer 1

`x y^2 y' = x + 1 \ \Rightarrow\ y^2 (dy)/(dx) = (x+1)/(x)\ \Rightarrow\ y^2 dy = \left(1 + (1)/(x)\right) dx\ \Rightarrow \\ \\ \int y^2 dy = \int \left(1 + (1)/(x)\right) dx\ \Rightarrow (1)/(3)y^3 = x + \ln|x| + C\ \Rightarrow \\ y^3 = 3x + 3\ln|x| + 3C\ \Rightarrow\ \\ \\y = \sqrt[3]x, \text{ where $K = 3C$}`