It is given that $2xy\frac{\text{d}y}{\text{d}x} = x^2 - y^2$ where $x > 0$, $y < 0$. Using the substitution $u = \frac{y}{x}$, show that the differential equation can be transformed to $\frac{2u}{1-3u^2}\frac{\text{d}u}{\text{d}x} = \frac{1}{x}$. Hence find the general solution of $y$ in terms of $x$.

1. Differentiate the substitution $y = ux$ with respect to $x$.

$y = ux \implies \frac{\text{d}y}{\text{d}x} = u + x\frac{\text{d}u}{\text{d}x}$

2. Substitute into the given differential equation and simplify to the target form.

$2x(ux)\left(u + x\frac{\text{d}u}{\text{d}x}\right) = x^2 - (ux)^2 \\ 2x^2u^2 + 2x^3u\frac{\text{d}u}{\text{d}x} = x^2(1 - u^2) \\ \text{Since } x > 0, \text{ divide by } x^2: \\ 2u^2 + 2xu\frac{\text{d}u}{\text{d}x} = 1 - u^2 \\ 2xu\frac{\text{d}u}{\text{d}x} = 1 - 3u^2 \\ \frac{2u}{1-3u^2}\frac{\text{d}u}{\text{d}x} = \frac{1}{x} \quad \text{(shown)}$

3. Separate variables and integrate both sides.

$\int \frac{2u}{1-3u^2} \, \text{d}u = \int \frac{1}{x} \, \text{d}x \\ -\frac{1}{3}\int \frac{-6u}{1-3u^2} \, \text{d}u = \int \frac{1}{x} \, \text{d}x \\ -\frac{1}{3}\ln|1-3u^2| = \ln x + C \quad (\text{since } x > 0, |x| = x)$

4. Rearrange to make $u^2$ the subject.

$\ln|1-3u^2| = -3\ln x - 3C \\ |1-3u^2| = e^{-3C}x^{-3} \\ 1-3u^2 = \frac{A}{x^3} \quad (\text{where } A = \pm e^{-3C}) \\ 3u^2 = 1 - \frac{A}{x^3} = \frac{x^3 - A}{x^3}$

5. Substitute back $u = \frac{y}{x}$ and solve for $y$, applying the condition $y < 0$.

$3\left(\frac{y}{x}\right)^2 = \frac{x^3 - A}{x^3} \\ \frac{3y^2}{x^2} = \frac{x^3 - A}{x^3} \\ y^2 = \frac{x^3 - A}{3x} \\ \text{Let } k = -A. \text{ Since } y < 0, \text{ we reject the positive root:} \\ y = -\sqrt{\frac{x^3 + k}{3x}}$