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i need to find all the solutions for this D.E. $$ y=xy'+(y')^3 $$

i did what seemed the correct move

$$y=y'(x+(y')^2)$$

but couldn't proceed more.

Lutz Lehmann
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Sam Peters
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    You might want to read this page: https://en.wikipedia.org/wiki/Clairaut's_equation. – Minus One-Twelfth Jun 09 '19 at 02:52
  • see also https://math.stackexchange.com/q/2739024/115115, https://math.stackexchange.com/q/2462239/115115, https://math.stackexchange.com/q/260349/115115 about the topic of Clairaut equations – Lutz Lehmann Jun 09 '19 at 11:46

2 Answers2

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This equation is called Clairut equation. Differentiate with respect to $x$ we have \begin{align*} y' &= y' + xy'' + 3(y')^2 y''\\ 0 &= (x + 3(y')^2)y'' \end{align*} If $x + 3(y')^2$ = 0 for all $x$ we have $(y') = \pm(x/3)^{1/2}$.

If $y'' = 0"$ we have $y = ax + b$ for some constant $a,b$. Plugging this to the equation we have $ b = a^3$. So the general solutions are given by $y = ax + a^3$ for any real number $a$.

Tom Himler
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Azlif
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  • You have a sign error in solving the condition for the singular solution. $x+3y'^2=0$ gives $x\le 0$ and $y'=\pm\sqrt{-x/3}$. – Lutz Lehmann Jun 09 '19 at 11:55
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Just from observation, if you assume a solution of the form $y=ax+b$, you get $y(x) = ax+a^3$ as a solution, where $a$ is any real number.

Edit: To find all solutions, differentiate the equation to obtain $$y'=y'+(x+3(y')^2)y''.$$ This equation is true iff $y''=0$ or $x+3(y')^2=0$. The first case corresponds to linear functions from my original answer. To find the rest of the solutions, we can combine $x+3(y')^2=0$ with the original differential equation to find a parametric formula for a curve in the $xy$ plane. If we denote $y'$ by a parameter $t$, then we have the following parametric equations: $$x=-3t^2,\\ y=xt+t^3.$$

The first can be inverted to obtain $t=\pm\sqrt{-x/3}$, which yields solutions $$y = \pm x\sqrt{\frac{-x}{3}}\pm\left(\frac{-x}{3}\right)^{3/2}, \ x\leq 0.$$ This can be written neater in terms of complex numbers.

Now these solutions are in fact all solutions to this problem because they were constructed by finding all solutions to the related second order equation then constraining them to solve the original first order equation

whpowell96
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    True. I will edit my answer – whpowell96 Jun 09 '19 at 02:38
  • There is no need to use complex numbers as one seeks a real solution. What you found is that there is only a solution for $x\le 0$ and then $y=t(x+t^2)=t(x-\frac13x)=\pm\frac23x\sqrt{\frac{-x}3}$. – Lutz Lehmann Jun 09 '19 at 10:43
  • The approach of "parameterized curve" is very nice. However, one should be careful in stating that these are the only solutions, because in fact they are not. They are the only branches where segments of solutions can lie, but true solutions can be combinations of these branches! – Luca Ghidelli Jun 28 '21 at 05:05