Solve the following differential equation: 2x²dy/dx - 2xy + y² = 0

Given: 2x²dy/dx - 2xy + y² = 0
First separate the terms:
2x²dy/dx = 2xy - y²
dy/dx = (2xy - y²) / 2x² → y/x - y²/2x²
dy/dx = y/x - (y/x)²
Let y = vx and dy/dx = v + x(dv/dx)
Then, v + x(dv/dx) = v - v²
x(dv/dx) = -v²
Now separate the variables, we get
-dv/v² = dx/x
Put the integral sign in front:
∫-1/v² dv = ∫1/x dx
2/v = log(x) + C
Now substitute back v = y/x
Therefore, 2x/y = log(x) + C