Verify that $x^2 + cy^2 = 1$ is an implicit solution to $\frac{dy}{dx} = \frac{xy}{x^2 - 1}.$

Question

When I differentiate $x^2 + cy^2 = 1$ implicitly and solve for $\dfrac{dy}{dx}$, I get $\dfrac{dy}{dx} = \dfrac{-x}{cy}.$ I thought I had to multiply by a fraction in order to make it similar to $\dfrac{dy}{dx} = \dfrac{xy}{x^2 - 1}$\$ but nothing worked. I then checked the answer and someone put up these steps: \begin{align*} y^2 &= -(x^2 - 1) * \dfrac{1}{c} \\ y^2 &= c(x^2 - 1) \\ y &= \sqrt{c(x^2 - 1)} \\ y &= c\sqrt{x^2 - 1)} \end{align*}

I don't get how $\dfrac{1}{c}$ changes to $c$ and it's square root disappears.

Answer 1

Notice that: $$ x^2 + cy^2 = 1 \iff cy = \frac{1 - x^2}{y} $$ So: $$ \frac{dy}{dx} = \frac{-x}{cy} = \frac{-x}{\left(\frac{1 - x^2}{y} \right)} = \frac{-xy}{1 - x^2} = \frac{xy}{x^2 - 1} $$

Answer 2

Note that $\frac{1}{c}$ is just some constant. Suppose I want to substitute $k=1/c$ This is what is actually happening here. They take note that $1/c$ is still just some constant and what they did was plug $c$ back in.

This same argument can be used on $\sqrt{c}$

Personally I don't like how the same $c$ was used over and over. How I would prefer to write it is this: $$y^2 = -(x^2+ 1)*\frac{1}{c_0}$$ Now substitute $c_1 = \frac{-1}{c_0}$ $$y^2 = c_1(x^2 +1)$$

Hopefully it makes more sense now, and I'll let you continue from here plugging in $c_2 = \sqrt{c_1}$