Solve $\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$ for $y$

Question

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 $y=-x$

Answer 1

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*}

Answer 2

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

Answer 3

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.

Answer 4

$$\frac{ky^2}{2}+cy=\frac{kx^2}{2}-cx \iff y^2+\frac{2cy}{k}=\frac2k \left (\frac{kx^2}{2}-ck \right ) \iff y^2+\frac{2cy}{k}+\frac{c^2}{k^2}=\frac2k \left (\frac{kx^2}{2}-ck \right )+\frac{c^2}{k^2} \iff \left (y+\frac{c}{k}\right )^2 =\frac{(c-kx)^2}{k^2}$$

I think you can continue from here.