Question

Solve the following differential equation:
$2x^2\frac{dy}{dx}-2xy+y^2=0.$

Answer 1

Given: $2x^2\frac{dy}{dx}-2xy+y^2=0$

First separate the terms:

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

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

$\frac{dy}{dx}=\frac{y}{x}-\frac{1}{2}{\left(\frac{y}{x}\right)}^2$

Let $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Then, $v+x\frac{dv}{dx}=v-\frac{v^2}{2}$

$x\frac{dv}{dx}=-\frac{v^2}{2}$

Now separate the variables, we get

$\frac{-2}{v^2}dv=\frac{1}{x}dx$

Put the integral sign in front:

$-2\int \frac{1}{v^2}dv=\int \frac{1}{x}dx$

$\frac{2}{v}=log\left(x\right)+C$

Now substitute back $v=\frac{y}{x}$

Therefore, $\frac{2x}{y}=log\left(x\right)+C$