Problem: I have to solve the following differential equation (which is supposed to be Clairaut's): $$y=x(y')^2-\frac{1}{y'},$$ where $y'=\frac{dy}{dx}.$

What I have tried: Because it says that is Clairaut's, I proceed by differentiating with respect to $x$. With an ideal CLairaut's differential equation, you could "isolate" every term that is multiplied by $\frac{d^2y}{dx}$ and, on the other side of the equality, you will get a zero. From there, it easy and you end up getting a family of curves [Link: Wikipedia]. The problem is the square in $x(y')^2,$ that messes up everything and I cannot isolate the terms in the right way.

I also tried using Mathematica

DSolve[{y[x] == x (y[x]')^2 - 1/(y[x]')}, y[x], x],

but it gives an error

$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of -(1/y').

I tried Python with Sympy, but failed too.

I tried a variable change, but I do not manage to find the optimal to simplify the equation and solve it following the regular procedure.

  • $\begingroup$ the Clairaut DE is of the form $y=xy'+...$ $\endgroup$ – Nosrati May 12 at 14:06
  • $\begingroup$ This is a d'Alembert's equation. $\endgroup$ – user10354138 May 12 at 14:14
  • $\begingroup$ Try to solve it as DSolve[y[x] y'[x] - x y'[x]^3 - 1 == 0, y, x] Mathematica gives a long answer. $\endgroup$ – Cesareo May 12 at 14:15
  • $\begingroup$ I couldn't solve it analytically, try to solve it with softwares. has it any initial condition? $\endgroup$ – Nosrati May 12 at 14:25

This is a d'Alembert's equation. Letting $p=y'$, we have $$ y=xp^2-\frac1p $$ Differentiating with respect to $x$, $$ p=p^2+2xp\frac{\mathrm{d}p}{\mathrm{d}x}+\frac1{p^2}\frac{\mathrm{d}p}{\mathrm{d}x} $$ So $$ \frac{\mathrm{d}x}{\mathrm{d}p}=\frac{2}{1-p}x+\frac1{p^3-p^4} $$ which integrates to $x = \frac{c}{(p-1)^2} - \frac1{2 (p - 1)^2 p^2} + \frac1{(p - 1)^2 p}$ and thus $$ y=xp^2-\frac1p=\dots $$

  • $\begingroup$ How do you integrate and get the $x=...$, because the first term on the right depends on $x$ and $p$. $\endgroup$ – plr May 12 at 20:53
  • $\begingroup$ You have $\frac{\mathrm{d}x}{\mathrm{d}p}+\frac{2}{p-1}x=\frac1{p^3-p^4}$ i.e., $\frac{\mathrm{d}}{\mathrm{d}p}((p-1)^2 x)=\frac{1-p}{p^3}$, etc. $\endgroup$ – user10354138 May 13 at 1:04

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