Answer to Question #138837 in Differential Equations for John Carl Mendevil

Question

dy/dx=2x+2xy²/y+2x²y

Answer 1

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

Let us solve this differential equation:

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

$\frac{ydy}{1+y^2}=\frac{2xdx}{1+2x^2}$ ($1+y^2>0$)

$\int\frac{ydy}{1+y^2}=\int\frac{2xdx}{1+2x^2}$

$\frac{1}{2}\int\frac{d(1+y^2)}{1+y^2}=\frac{1}{2}\int\frac{d(1+2x^2)}{1+2x^2}$

$\int\frac{d(1+y^2)}{1+y^2}=\int\frac{d(1+2x^2)}{1+2x^2}$

$\ln(1+y^2)=\ln(1+2x^2)+ln(C)$

$\ln(1+y^2)=\ln((1+2x^2)C)$

Therefore, the general solution is the following:

$1+y^2=C(1+2x^2)$