# Clairuts' Equation

This a tough one it seems:

$$(xp-y)^2=a(1+p^2)(x^2+y^2)^{3/2}$$

where $p = dy/dx$.

I tried using $x = r\cos a$ and $y=r\sin a$ but it just keeps getting more complicated than simplifying. Help?

Also can all equations of first order but not of first degree be converted to Clairut's form? THANKS!

• its actually supposed to b sin a , you're right about p . The substitution doesn't really work though :/ – user113051 Dec 10 '13 at 10:19
• Is $a$ a constant? or is it a free variable? – Willie Wong Dec 10 '13 at 10:28
• a is a constant. really sorry on all the detail . One could think of making it solvable for p but that I tried and it din work :/ – user113051 Dec 10 '13 at 10:30
• Clairaut.  – Did Jun 28 '14 at 22:35

$\displaystyle\left(x\dfrac{\mathrm dy}{\mathrm dx}-y\right)^2=a\left(1+\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2\right)(x^2+y^2)^{3/2}$
$\displaystyle \begin{cases}x=\rho \cos(\theta) \\ y=\rho \sin(\theta)\end{cases}\,\to\,\begin{cases}\mathrm dx=\mathrm d\rho\cos(\theta)-\rho\sin(\theta)\mathrm d\theta \\ \mathrm dy=\mathrm d\rho\sin(\theta)+\rho\cos(\theta)\mathrm d\theta\end{cases}$
$\displaystyle\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\dfrac{\mathrm d\rho}{\mathrm d\theta}\sin(\theta)+\rho\cos(\theta)}{\dfrac{\mathrm d\rho}{\mathrm d\theta}\cos(\theta)-\rho\sin(\theta)}$
$\displaystyle\left( \rho \cos(\theta) \dfrac{\dfrac{\mathrm d\rho}{\mathrm d\theta}\sin(\theta)+\rho\cos(\theta)}{\dfrac{\mathrm d\rho}{\mathrm d\theta}\cos(\theta)-\rho\sin(\theta)}-\rho\sin(\theta) \right)^2=a\left[1+\left(\dfrac{\dfrac{\mathrm d\rho}{\mathrm d\theta}\sin(\theta)+\rho\cos(\theta)}{\dfrac{\mathrm d\rho}{\mathrm d\theta}\cos(\theta)-\rho\sin(\theta)}\right)^2\right]\rho^3$