<s>[INST] Solve first order partial diferential equation Consider the Cauchy problem:
$$\left\{\begin{array}{lll}
x^2\partial_x u+y^2\partial_yu=u^2\\
u(x,2x)=1
\end{array}\right.$$
It is easy to show that the characteristic equations are given by:
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{px^2+qy^2}=\frac{dp}{2pu-2xp}=\frac{dq}{2qu-2yq}=dt$$
By the Lagrange-Charpit's method, we need a first integral $\Psi$ different to $\Phi=x^2p+y^2q-u^2$.
How I can find  $\Psi$? Maybe we should get directly the characteristic curves of the previous system?
Many thanks!
 [/INST] Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x^2$ , letting $x(1)=-1$ , we have $-\dfrac{1}{x}=t$
$\dfrac{dy}{dt}=y^2$ , we have $-\dfrac{1}{y}=t+y_0=-\dfrac{1}{x}+y_0$
$\dfrac{du}{dt}=u^2$ , we have $\dfrac{1}{u}=-t+f(y_0)=\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$ , i.e. $u(x,y)=\dfrac{1}{\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)}$
$u(x,2x)=1$ :
$\dfrac{1}{\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{2x}\right)}=1$
$\dfrac{1}{x}+f\left(\dfrac{1}{2x}\right)=1$
$f\left(\dfrac{1}{2x}\right)=1-\dfrac{1}{x}$
$f(x)=1-\dfrac{1}{\dfrac{1}{2x}}=1-2x$
$\therefore u(x,y)=\dfrac{1}{\dfrac{1}{x}+1-2\left(\dfrac{1}{x}-\dfrac{1}{y}\right)}=\dfrac{1}{1+\dfrac{2}{y}-\dfrac{1}{x}}=\dfrac{xy}{xy+2x-y}$
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