An odd equation that is a straight line: $x(y^2-x^2)=y^3$

Question

I was idly messing around in Desmos the other day and came up with this equation:

$$x(y^2-x^2)=y^3$$

The graph looks like this. A perfectly straight line.

My first instinct was to differentiate it, but that didn't help much. Because it's implicit, the instantaneous slope requires both an $x$ and a $y$, and I don't have one in terms of the other.

Asking WolframAlpha to solve the equation in terms of $y$ gives me a slope of

$m=-\frac1 6(-2+2^{2/3}\sqrt[3]{25-3\sqrt{69}}+2^{2/3}\sqrt[3]{25+3\sqrt{69}})\approx-0.75488$

Why and how does this happen? The equation does not look like it should be linear in any way haha. Sorry if my relatively low level of knowledge makes it difficult to explain

Answer 1

You can re-arrange the terms to get $$x^3 - x y^2 + y^3 = 0$$ and you see that all the terms added have a combined power of 3. It is reasonable to proceed as follows:

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Indeed, the equation $y = \alpha x$ is a solution, for a specific value of $\alpha$.

Usee the above in the equation, to get

$$ x \left( (\alpha x)^2 - x^2 \right) = \alpha^3 x^3 $$

and move all terms to the right side and collect the $x^3$ terms

$$ x^3 \left( \alpha^3 - \alpha^2 + 1 \right) = 0 $$

As you can see when $\alpha^3 - \alpha^2 + 1=0$ then $y= \alpha x$ is a solution.

The solution for $\alpha$ has only one real value

$$ \alpha = \tfrac{1}{3} - \sqrt[3]{\tfrac{25}{54} - \tfrac{\sqrt{69}}{18}} - \sqrt[3]{ \tfrac{25}{54} + \tfrac{\sqrt{69}}{18}} \approx -0.7548776662466927 $$

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You can also do the implicit differentiation to get

$$ \frac{{\rm d}y}{{\rm d}x} = \frac{3 x^2-y^2}{y(2x - 3 y)} $$

and again, based on the graph, see if $y = \alpha x$ solves the above. Note that the slope $\alpha = \tfrac{{\rm d}y}{{\rm d}x}$ results in

$$ \alpha = \frac{3 x^2 - \alpha^2 x^2}{\alpha x ( 2 x - 3 \alpha x)} = \frac{\alpha^2-3}{\alpha ( 3 \alpha -2)}$$

with the same solution for $\alpha$ as before.

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More interestingly if you ploted $x^3 - 3 x y^2 + y^3 = 0$ you would see three lines that meet at the origin.

This is because the solution of $\alpha^3 -3 \alpha^2 + 1=0$ has three real roots.

Answer 2

The real solution of the cubic equation $$y^3-x\,y^2+x^3=0$$ is $$y=\frac{1}{3} \left(1-2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{25}{2}\right)\right)\right)\,x$$

Answer 3

You can factorise the expression $$ y^3-xy^2+x^3 $$ into a product of a linear and quadratic term: $$ y^3-xy^2+x^3=(y-m x)\big(y^2-(1-m)yx-m^{-1}x^2\big)\ , $$ where $\ m\ $ is the (unique) real root of the cubic polynomial $$ m^3-m^2+1\ , $$ the value of which is just the slope of the line Wolfram alpha returned in response to your request.

You can write the quadratic factor as $$ y^2-(1-m)yx-m^{-1}x^2=\pmatrix{y&x}\pmatrix{1&-\frac{1-m}{2}\\-\frac{1-m}{2}&\frac{-1}{m}}\pmatrix{y\\x}\ . $$ The matrix $$ \pmatrix{1&-\frac{1-m}{2}\\-\frac{1-m}{2}&\frac{-1}{m}} $$ of this quadratic form has eigenvalues $\ \frac{m-1+\sqrt{2+m^2+m}}{2m}\ $ and $\ \frac{m-1-\sqrt{2+m^2+m}}{2m}\ $, both of which are positive. The matrix is therefore positive definite, and the quadratic form $\ y^2-(1-m)yx-m^{-1}x^2\ $ is strictly positive for all real $\ x\ $ and $\ y\ $ except $\ x=y=0\ $.

Thus, the only way the equation \begin{align} 0&=y^3-xy^2+x^3\\ &=(y-m x)\big(y^2-(1-m)yx-m^{-1}x^2\big) \end{align} can be satisfied is if $\ y=mx\ $.