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This differential equation showed up in a geometry problem

$$ \left(x - \frac{y}{y'}\right)^2 \left(1 + \left(y'\right)^2\right) = 1 $$

I figured out by trial and error that $y(x) = \left( 1 - x^{2/3} \right)^{3/2}$, the graph of an astroid, is a solution, but I'd like to see a way how this can be solved in a more proper manner, rather than just by getting lucky, and if multiple solutions exist.

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  • $\begingroup$ What classes of easily solved DE do you know that allow complicated expressions in $y'$? $\endgroup$ Aug 24, 2023 at 14:05
  • $\begingroup$ I'm not sure. Separable DEs? $\endgroup$
    – Kasper
    Aug 24, 2023 at 14:21
  • $\begingroup$ Could you verify that $y(x)=(1-x^{2/3})^{3/2}$ is indeed a solution? If I substitute it into your ODE, I get $x^{-1/3}\neq 1$. $\endgroup$
    – CW279
    Aug 24, 2023 at 14:29
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    $\begingroup$ @CW279 I forgot to square the first factor, it should be correct now $\endgroup$
    – Kasper
    Aug 24, 2023 at 14:33
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    $\begingroup$ Then what class do you suggest? $\endgroup$
    – Kasper
    Aug 24, 2023 at 15:03

3 Answers 3

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This can be transformed into the normal form of a Clairaut equation $y=xy'+f(y')$, or two Clairaut equations to account for the square root $$ y=xy'\pm\frac{y'}{\sqrt{1+y'^2}} $$ The regular solutions are the linear functions $$ y=Cx\pm\frac{C}{\sqrt{1+C^2}} $$ Additionally there are the singular solutions which solve $x+f'(y')=0$, here $$ x\pm\frac{1}{\sqrt{1+y'^2}^3}=0. $$

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Hint.

Calling $p=y'$ we have

$$ \left(x-\frac yp\right)^2(1-p^2)-1=0\ \ \ \ \ \ (1) $$

now deriving

$$ 2 \left(1-p^2\right) \left(x-\frac{y}{p}\right) \left(\frac{y p'}{p^2}-\frac{y'}{p}+1\right)-2 p \left(x-\frac{y}{p}\right)^2 p'=0 $$

or

$$ 2 \left(1-p^2\right) \left(x-\frac{y}{p}\right) \frac{y p'}{p^2}-2 p \left(x-\frac{y}{p}\right)^2 p'=0\ \ (2) $$

now eliminating $y$ between $(1)$ and $(2)$ we get at

$$ \cases{ p'=0\\ x^2 p^2 \left(p^4-3 p^2+3\right)-x^2+1=0\ \ \ \ (3) } $$

NOTE

Solving $(3)$ for $p$, one of the folds gives us

$$ p = y' = \sqrt{1-\frac{1}{x^{2/3}}}\Rightarrow y = \left(1-\frac{1}{x^{2/3}}\right)^{3/2} x $$

as a particular solution and from $p'=0$ we obtain

$$ y''=0\Rightarrow y = c_1 x + c_2 $$

such that

$$ \left(\frac{1}{c_1^2}-1\right) c_2^2-1=0 $$

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The command of Maple 2023

[dsolve((x - y(x)/diff(y(x), x))^2*(1 + diff(y(x), x)^2) = 1)];

produces $$y\! \left(x\right) {=} \sqrt{3 x^{\frac{4}{3}}-3 x^{\frac{2}{3}}-x^{2}+1},y\! \left(x\right){=} -\sqrt{3 x^{\frac{4}{3}}-3 x^{\frac{2}{3}}-x^{2}+1}, \\ y\! \left(x\right){=}-\frac{\sqrt{4-6 \,\mathrm{I} x^{\frac{4}{3}} \sqrt{3}-6 \,\mathrm{I} x^{\frac{2}{3}} \sqrt{3}-6 x^{\frac{4}{3}}+6 x^{\frac{2}{3}}-4 x^{2}}}{2}, \\ y\! \left(x\right){=}\frac{\sqrt{4-6 \,\mathrm{I} x^{\frac{4}{3}} \sqrt{3}-6 \,\mathrm{I} x^{\frac{2}{3}} \sqrt{3}-6 x^{\frac{4}{3}}+6 x^{\frac{2}{3}}-4 x^{2}}}{2}, \\ y\! \left(x\right){=}-\frac{\sqrt{4+6 \,\mathrm{I} x^{\frac{4}{3}} \sqrt{3}+6 \,\mathrm{I} x^{\frac{2}{3}} \sqrt{3}-6 x^{\frac{4}{3}}+6 x^{\frac{2}{3}}-4 x^{2}}}{2}, \\ y\! \left(x\right)l{=}\frac{\sqrt{4+6 \,\mathrm{I} x^{\frac{4}{3}} \sqrt{3}+6 \,\mathrm{I} x^{\frac{2}{3}} \sqrt{3}-6 x^{\frac{4}{3}}+6 x^{\frac{2}{3}}-4 x^{2}}}{2}, \\ y\! \left(x\right){=}x c_{1}-\frac{c_{1}}{\sqrt{c_{1}^{2}+1}},y\! \left(x\right) {=} x c_{1}+\frac{c_{1}}{\sqrt{c_{1}^{2}+1}}.$$

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