Solve $\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$ for $y$ I have the equation:
$\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$,
where $k$ and $c$ are arbitrary constants.
How do I go about simplifying this and solving for $y$ in terms of $x$, excluding the obvious solution $y=-x$

HINT: Let $D$ be left hand side of the equation. Then $\frac12ky^2+cy=D$ is a quadratic equation, which we can solve.
