Derivative of xy on the same side

Question

How do you derive the solution when there is $xy$ on the same side of the equation?

For example, $$\frac{dy}{dx} = \frac{6xy}{3-x^2} $$

I have tried solving for $y$ or am I to find $dy?$

Answer 1

$$\frac{dy}{dx}=\frac{6xy}{3-x^2}\implies \frac{dy}{y}=\frac{6x}{3-x^2}dx\implies \int\frac{dy}{y}=\int\frac{6x}{3-x^2}dx\implies \ln(|y|)=-3\ln(|3-x^2|)+C\implies |y|=|3-x^2|^{-3}\implies y=\pm|3-x^2|^{-3}+C$$

Answer 2

Separating the variables and integrating, we get $$ \int\frac{\mathrm{d}y}{y}=\int\frac{6x\,\mathrm{d}x}{3-x^2} $$ which means $$ \log(y)=-3\log(3-x^2)+\log(C) $$ or $$ y(3-x^2)^3=C $$ Since the integrals don't converge at $y=0$ or $x^2=3$, solution domains do not contain those points. A different $C$ can be used in each of the six regions where $y\lt0$ and $y\gt0$ and where $x\lt-\sqrt3$ and $-\sqrt3\lt x\lt\sqrt3$ and $x\gt\sqrt3$.

Thus, to get $y$ as a function of $x$, we get $$ y=\frac{C}{(3-x^2)^3} $$ where a different $C$ can be used in each of the three regions $x\lt-\sqrt3$ and $-\sqrt3\lt x\lt\sqrt3$ and $x\gt\sqrt3$ since the solution cannot be continuous across the boundaries between them.