Answer to Question #300203 in Differential Equations for Joshua

Question

Solve the Bernoulli Equation $xyy'+y^2=2x$

Answer 1

Given:

$xyy'+y^2=2x$

We use the substitution

$$v=y^2$$

Then

$$v'=2yy'$$

We get the next DE

$$\frac{x}{2}v'+v=2x$$

$$v'+\frac{2}{x}v=4$$

$$\mu(x)=e^{\int \frac{2}{x}dx}=e^{2 \ln x}=x^2$$

$$x^2v'+2xv=4x^2$$

$$(x^2v)'=4x^2$$

$$x^2v=(4/3)x^3+C$$

$$v=(4/3)x+C/x^2$$

$$y=\pm\sqrt{(4/3)x+C/x^2}$$