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 $y=-x$
 A: Treat $x$ as a constant as well. We could use the standard quadratic formula, but since you noticed that $y = -x$ is a solution, let's try factoring instead. We obtain:
\begin{align*}
0
&= \tfrac{1}{2}ky^{2} - \tfrac{1}{2}kx^{2} + cy + cx \\
&= \tfrac{1}{2}k(y^{2} - x^{2}) + c(y + x) \\
&= \tfrac{1}{2}k(y - x)(y + x) + c(y + x) \\
&= (\tfrac{1}{2}k(y - x) + c)(y + x) \\
\end{align*}
Thus, the other solution can be obtained by setting the first factor equal to zero, yielding:
\begin{align*}
\tfrac{1}{2}k(y - x) + c &= 0 \\
\tfrac{1}{2}k(y - x) &= -c \\
y - x &= \tfrac{-2c}{k} \\
y &= x - \tfrac{2c}{k}
\end{align*}
A: HINT: Let $D$ be left hand side of the equation. Then $\frac12ky^2+cy=D$ is a quadratic equation, which we can solve.
A: your equation can be factorized as follows $-1/2\, \left( x+y \right)  \left( -kx+ky+2\,c \right) =0$ from here you will get all solutions.
A: $$\frac{ky^2}{2}+cy=\frac{kx^2}{2}-cx \iff y^2+\frac{2cy}{k}=\frac2k \left (\frac{kx^2}{2}-ck \right ) \iff y^2+\frac{2cy}{k}+\frac{c^2}{k^2}=\frac2k \left (\frac{kx^2}{2}-ck \right )+\frac{c^2}{k^2} \iff \left (y+\frac{c}{k}\right )^2 =\frac{(c-kx)^2}{k^2}$$
I think you can continue from here.
