We are given a first-order differential equation and we are asked to solve it. The equation is $y' = \frac{y^2}{xy - x^2}$.

1. Problem Description

We are given a first-order differential equation and we are asked to solve it. The equation is $y' = \frac{y^2}{xy - x^2}$.

2. Solution Steps

The given differential equation is $y' = \frac{y^2}{xy - x^2}$.

We can rewrite this as $y' = \frac{y^2}{x(y - x)}$. This is a homogeneous differential equation, so we can use the substitution $y = vx$, where $v$ is a function of $x$.

Then, $y' = v + xv'$. Substituting these into the equation, we have

$v + xv' = \frac{(vx)^2}{x(vx - x)} = \frac{v^2x^2}{x^2(v - 1)} = \frac{v^2}{v - 1}$.

Then, $xv' = \frac{v^2}{v - 1} - v = \frac{v^2 - v(v - 1)}{v - 1} = \frac{v^2 - v^2 + v}{v - 1} = \frac{v}{v - 1}$.

Separating variables, we have $\frac{v - 1}{v}dv = \frac{dx}{x}$.

Integrating both sides, we get

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

$\int (1 - \frac{1}{v})dv = \int \frac{dx}{x}$

$v - \ln|v| = \ln|x| + C$, where $C$ is the constant of integration.

Now, we substitute $v = \frac{y}{x}$.

$\frac{y}{x} - \ln|\frac{y}{x}| = \ln|x| + C$

$\frac{y}{x} - \ln|y| + \ln|x| = \ln|x| + C$

$\frac{y}{x} - \ln|y| = C$

$\frac{y}{x} = \ln|y| + C$

$y = x(\ln|y| + C)$

3. Final Answer

$y = x(\ln|y| + C)$

$\frac{y}{x} - \ln|y| = C$