How to find general solution of a differential equation with two first-order terms?

Question

Find the GS of the DE:

$y'=2(xy'+y)y^3$

Answer 1

$$y'=2(xy'+y)y^3$$ Rearranging terms $(y \neq 0, y =0 \text { is also a solution to the ode})$ $$\frac {y'}{y^3}=2(xy)'$$ Simply integrate : $$\int \frac {y'}{y^3}dx=2\int(xy)'dx$$ $$\color{blue}{ \int \frac {y'}{y^3}dx=\int \frac 1 {y^3} \frac {dy}{dx}dx=\int \frac {dy}{y^3}=-\frac {1}{2y^2}}$$ $$-\frac 12 \left (\frac {1}{y^2} \right )=2(xy)+K$$ $$\frac {1}{y^2}=-4xy+K$$ You leave it this way or you express x as a function of y $$Ky^2-4xy^3=1$$ X as a function of y: $$\boxed {x=\frac Ky-\frac 1 {4y^3}}$$

Answer 2

Another way to solve this (this would work in a more general case).

$$y'=2x y^3 y'+2 y^4$$

Assume $y' \neq 0$, then:

$$1=2x y^3+2 y^4 \frac{1}{y'}$$

Since $$\frac{1}{y'}=\frac{dx}{dy}=x'$$

We can rewrite ($y \neq 0$):

$$y x'+x=\frac{1}{2y^3}$$

Solving homogeneous equation first:

$$yx_h'+x_h=0$$

$$x_h=\frac{C}{y}$$

Now use the method of variable constants for the particular solution:

$$x_p=\frac{C(y)}{y}$$

$$C'-\frac{C}{y}+\frac{C}{y}=\frac{1}{2y^3}$$

$$C'=\frac{1}{2y^3}$$

$$C(y)=-\frac{1}{4 y^2}$$

Finally:

$$x(y)=\frac{C}{y}-\frac{1}{4 y^3}$$