# Solution to differential equation $(x - y/y')^2 (1 + (y')^2) = 1$

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.

• What classes of easily solved DE do you know that allow complicated expressions in $y'$? Aug 24, 2023 at 14:05
• I'm not sure. Separable DEs? Aug 24, 2023 at 14:21
• 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 $x^{-1/3}\neq 1$. Aug 24, 2023 at 14:29
• @CW279 I forgot to square the first factor, it should be correct now Aug 24, 2023 at 14:33
• Then what class do you suggest? Aug 24, 2023 at 15:03

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 $$x\pm\frac{1}{\sqrt{1+y'^2}^3}=0.$$

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

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

The command of Maple 2023

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


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