# A differential equation (nonlinear First-Order)

how to solve this equation:

$(Px-y)(Py+x)=h^2P$

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

$\therefore\left(\dfrac{x}{2y}\left(\dfrac{du}{dx}-2x\right)-y\right)\dfrac{1}{2}\dfrac{du}{dx}=\dfrac{h^2}{2y}\left(\dfrac{du}{dx}-2x\right)$

$\left(x\left(\dfrac{du}{dx}-2x\right)-2y^2\right)\dfrac{du}{dx}=2h^2\left(\dfrac{du}{dx}-2x\right)$

$x\dfrac{du}{dx}-2x^2-2y^2=2h^2\left(1-\dfrac{2x}{\dfrac{du}{dx}}\right)$

$x\dfrac{du}{dx}-2u=2h^2-\dfrac{4h^2x}{\dfrac{du}{dx}}$

$2u+2h^2=x\dfrac{du}{dx}+\dfrac{4h^2x}{\dfrac{du}{dx}}$

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

Let $v=x^2$ ,

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

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

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

Let $s=u+h^2$ ,

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

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

$s\dfrac{dv}{ds}=v+\dfrac{h^2}{\left(\dfrac{ds}{dv}\right)^2}$

$v=s\dfrac{dv}{ds}-h^2\left(\dfrac{dv}{ds}\right)^2$

Which reduces to Clairaut's ODE.

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

$\dfrac{d^2v}{ds^2}\left(2h^2\dfrac{dv}{ds}-s\right)=0$

$\therefore\begin{cases}\dfrac{d^2v}{ds^2}=0\\2h^2\dfrac{dv}{ds}-s=0\end{cases}$

$\begin{cases}v=as+b\\v=\dfrac{s^2}{4h^2}+c\end{cases}$

$\therefore\begin{cases}as+b=as-h^2a^2\\\dfrac{s^2}{4h^2}+c=\dfrac{s^2}{2h^2}-\dfrac{s^2}{4h^2}\end{cases}$

$\begin{cases}b=-h^2a^2\\c=0\end{cases}$

$\therefore\begin{cases}v=as-h^2a^2\\v=\dfrac{s^2}{4h^2}\end{cases}$

$\begin{cases}x^2=au+h^2a-h^2a^2\\x^2=\dfrac{(u+h^2)^2}{4h^2}\end{cases}$

$\begin{cases}x^2=ax^2+ay^2+h^2a-h^2a^2\\x^2=\dfrac{(x^2+y^2+h^2)^2}{4h^2}\end{cases}$

$\begin{cases}(a-1)x^2+ay^2=h^2a^2-h^2a\\x^2=\dfrac{(x^2+y^2+h^2)^2}{4h^2}\end{cases}$

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