1. ## Clairaut's Equation

I'm starting on non-linear differential equations and there was a particular example in the book which didn't make much sense to me. It is as follows:

Solve the differential equation
$\displaystyle \left( x^{2}\; -\; 1 \right)p^{2}\; -\; 2xyp\; +\; y^{2}\; -\; 1\; =\; 0$
They rewrite the equation in this form:
$\displaystyle \left( y\; -\; xp \right)^{2}\; -\; 1\; -\; p^{2}\; =\; 0$

And the say that it "could be broken into two equations, each of Clairaut's form." This assumption confuses me. What is the logic behind it?

oh yeah! $\displaystyle p\; =\; \frac{dy}{dx}$

2. Originally Posted by jameselmore91
I'm starting on non-linear differential equations and there was a particular example in the book which didn't make much sense to me. It is as follows:

Solve the differential equation
$\displaystyle \left( x^{2}\; -\; 1 \right)p^{2}\; -\; 2xyp\; +\; y^{2}\; -\; 1\; =\; 0$
They rewrite the equation in this form:
$\displaystyle \left( y\; -\; xp \right)^{2}\; -\; 1\; -\; p^{2}\; =\; 0$

And the say that it "could be broken into two equations, each of Clairaut's form." This assumption confuses me. What is the logic behind it?

oh yeah! $\displaystyle p\; =\; \frac{dy}{dx}$
Dear ameselmore91,

$\displaystyle \left( y\; -\; xp \right)^{2}\; -\; 1\; -\; p^{2}\; =\; 0$

$\displaystyle (y-xp)^2=1+p^2$

$\displaystyle (y-xp)=\pm\sqrt{1+p^2}$

$\displaystyle y=xp\pm\sqrt{1+p^2}$

This equation is of the Clairaut's form, hence the general solution is,

$\displaystyle y=cx\pm\sqrt{1+c^2}~;~where~c~is~an~arbitary~const ant.$