To Solve $\displaystyle q+xp=p^2$ using Charpit's Method

Question

p = ∂z/∂x
q = ∂z/∂y

My Attempt:

$\displaystyle \frac{dx}{-(x-2p)}=\frac{dy}{-1}=\frac{dz}{-p(x-2p)-q}=\frac{dp}{2p}=\frac{dq}{0}$

So, do I get.. $\displaystyle \log p=\log c_1 \implies p=c_1 $

$\displaystyle q=c_2 $

Going to $\displaystyle dz=pdx+qdy $

This won't give the desired solution

Now, I can't think ahead ..

The given answer is: $\displaystyle z=axe^{-y}-\frac{1}{2}a^2e^{-2y}+b$

Answer 1

$\dfrac{\partial z}{\partial y}+x\dfrac{\partial z}{\partial x}=\left(\dfrac{\partial z}{\partial x}\right)^2$

$\dfrac{\partial^2z}{\partial x\partial y}+x\dfrac{\partial^2z}{\partial x^2}+\dfrac{\partial z}{\partial x}=2\dfrac{\partial z}{\partial x}\dfrac{\partial^2z}{\partial x^2}$

$\dfrac{\partial^2z}{\partial x\partial y}+\left(x-2\dfrac{\partial z}{\partial x}\right)\dfrac{\partial^2z}{\partial x^2}=-\dfrac{\partial z}{\partial x}$

Let $u=\dfrac{\partial z}{\partial x}$ ,

Then $\dfrac{\partial u}{\partial y}+(x-2u)\dfrac{\partial u}{\partial x}=-u$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$

$\dfrac{du}{dt}=-u$ , letting $u(0)=u_0$ , we have $u=u_0e^{-t}=u_0e^{-y}$

$\dfrac{dx}{dt}=x-2u=x-2u_0e^{-t}$ , we have $x=u_0e^{-t}+f(u_0)e^t=u+f(ue^y)e^y$

Answer 2

The Lagrange-Charpit equations have some small error in the $p$ component, the factor $2$, as with $f=p^2-px-q$ one has $f_x+pf_z=-p$.

The easy relations are $q=q_0=const.$ and $-y=\ln|p|+C$ or $p=ae^{-y}$.

Using the original equation $q=q_0=a^2e^{-2y}-axe^{-y}$ describes the characteristic curves.

Using $−p(x−2p)−q=p^2=a^2e^{2y}$ the $z$ and $p$ fractions combine to $dz=p\,dp$ which integrates to $$z=\frac12p^2+b=\frac12a^2e^{-2y}+b$$ This can be modified by adding the identity $0=axe^{-y}-a^2e^{-2y}+q_0$ to obtain the given solution.