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how to solve this equation:


that $P=\frac{dy}{dx}$

and $h$ is a constant.

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Typically, one tries to find an integrating factor... Your equation looks a bit suspicious. Is it homework/exercise in a book? Did you copy the problem correctly? As it is stated it is not a second order DE, but a (quite nonlinear) first order DE. – Fabian May 3 '11 at 7:12
this is a homework,i say "I know it gets First Order" because our lesson treat on FO DE :D,its equal to : $xyP^2+(x^2-y^2-h^2)P-xy=0$, can help? – Doman May 3 '11 at 7:23
It's a first order ODE, since it involves $y(x)$ and the first derivative $y'(x)$. (Not higher derivatives like $y''(x)$ etc.) What's your question, exactly? – Hans Lundmark May 3 '11 at 7:42
@Doman: even when there is $P^2$ it is still a first order ODE (no $y''(x)$ appearing). It is just a nonlinear differential equation. – Fabian May 3 '11 at 7:55
@All:thanks, what a bad mistake :D. Q EDITED. – Doman May 3 '11 at 14:01

Assume $h\neq0$ for the key case:

Let $u=x^2+y^2$ ,

Then $\dfrac{du}{dx}=2x+2y\dfrac{dy}{dx}$







Let $v=x^2$ ,

Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=2x\dfrac{du}{dv}$

$\therefore u+h^2=x^2\dfrac{du}{dv}+\dfrac{h^2}{\dfrac{du}{dv}}$


Let $s=u+h^2$ ,

Then $\dfrac{ds}{dv}=\dfrac{du}{dv}$

$\therefore s=v\dfrac{ds}{dv}+\dfrac{h^2}{\dfrac{ds}{dv}}$



Which reduces to Clairaut's ODE.











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