1
$\begingroup$

Can anyone please help me to find the solution of:

$$(y')^2-2xy'+y=0$$

using parametrization.

I know that it is d'Alembert's equation, and I found two family of solutions

$$x=\frac23(x+(x^2-y)^{1/2})+\frac c{(x+(x^2-y)^{1/2})^2}$$ and

$$x=\frac23(x-(x^2-y)^{1/2})+\frac c{(x-(x^2-y)^{1/2})^2}$$

but I need to solve it by parametrization.

The instructions are: "solve the equation by exhibiting parametrization for each of its integral curves."

Any ideas or solutions will be greatly appreciate it.

Also, I need to find it's singular curve, which I think is the line y=0. Please correct me if I'm wrong.

Thank you.

$\endgroup$

1 Answer 1

1
$\begingroup$

With $p=y'$ as parameter where $y'$ is not a constant, $\dot x=\frac{dx}{dp}$, $\dot y=\frac{dy}{dp}$ one gets $$ y-2px+p^2=0~\implies~ p\dot x=\dot y=2x+2p\dot x-2p,\\ $$ Solving the resulting differential equation using $p$ as integrating factor $$ 0=2px+p^2\dot x-2p^2~\implies~ c=p^2x-\frac23p^3 $$ Now this can be solved for $x$ and inserted into the original equation to find also $y$ as function of $p$. \begin{align} x&=\frac23p+\frac{c}{p^2}\\ y&=\frac13p^2+\frac{2c}p \end{align}


Solving the original equation as quadratic equation in $p$ indeed results in $(p-x)^2=x^2-y$, $p=x\pm\sqrt{x^2-y}$. For the square one gets $p^2=2px-y=2x^2-y\pm2x\sqrt{x^2-y}$. Inserted this gives $$ x=\frac23\left(x\pm\sqrt{x^2-y}\right)+\frac{c}{2x^2-y\pm2x\sqrt{x^2-y}} $$ which looks different to your implicit solutions.

$\endgroup$
3
  • $\begingroup$ Thank you for your answer Lutzl. Can you please elaborate how did you get your first two implications? $\endgroup$
    – Oli
    Mar 28, 2018 at 13:22
  • $\begingroup$ You have, by some variant of the chain rule, $\dot y=\frac{dy}{dp}=\frac{dy}{dx}\frac{dx}{dp}=p\dot x$. For the other side take the derivative by $p$ of $y=2px-y^2$. $\endgroup$ Mar 28, 2018 at 13:26
  • $\begingroup$ Thank you, I understood. I also edited the solutions in my question. I forgot to add the square. Now, they match your solutions. I appreciate your help. $\endgroup$
    – Oli
    Mar 28, 2018 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.