# Finding solution of differential equation with parametrization

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.

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.
• 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 $y=2px-y^2$. Mar 28, 2018 at 13:26