I have a partial differential equation $$u_xu_y = xy \mbox{ with } u(0,y)=y+1$$ Calling $u_x = p, u_y = q$ gives the following Charpit equations
$$\frac{dx}{dt} = q, \frac{dy}{dt} = p, \frac{dp}{dt} = y, \frac{dq}{dt} = x, \frac{du}{dt} = 2pq$$
How must I continue?
In $t=0$:
$x=0$, $y=s$, $u=s+1$ and $\frac{du}{ds}=\frac{du}{dx}\frac{dx}{ds}+\frac{du}{dy}\frac{dy}{ds}$ so that $1=p\cdot 0+q\cdot 1$ so $q=1$. Also: $pq=xy$ so that $p=0$.
Continue with $x''=x$ and $y''=y$:
$x(t)=c_1 e^{t}+c_2 e^{-t}$ and $y(t)=c_3 e^{t}+c_4 e^{-t}$ . So $q(t)=c_1 e^{t}-c_2 e^{-t}$ and $p(t)=c_3 e^{t}+c_4 e^{-t}$. Initial conditions give $c_1=\frac{1}{2},c_2=-\frac{1}{2},c_3=c_4=\frac{1}{2}s$.
Because $u'=2pq=2s\cdot \sinh(t)\cosh(t)=s\cdot \sinh(2t)$ so that $u(t)=\frac{s}{2}\cdot \cosh(2t)+c$. Using the initial condition $u(0)=s+1$ gives $c=s+1$.
Now it's up to you to write this in more explicit form. What is $x$ and what is $y$?
Let $\begin{cases}p=x^2\\q=y^2\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=2x\dfrac{\partial u}{\partial p}$
$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial y}=2y\dfrac{\partial u}{\partial q}$
$\therefore2xu_p2yu_q=xy$ with $u(0,q)=\pm\sqrt q+1$
$u_pu_q=\dfrac{1}{4}$ with $u(0,q)=\pm\sqrt q+1$
$u_p-\dfrac{1}{4u_q}=0$ with $u(0,q)=\pm\sqrt q+1$
$u_{pq}+\dfrac{u_{qq}}{4u_q^2}=0$ with $u(0,q)=\pm\sqrt q+1$
Let $v=u_q$ ,
Then $v_p+\dfrac{v_q}{4v^2}=0$ with $v(0,q)=\pm\dfrac{1}{2\sqrt q}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dp}{dt}=1$ , letting $p(0)=0$ , we have $p=t$
$\dfrac{dv}{dt}=0$ , letting $v(0)=v_0$ , we have $v=v_0$
$\dfrac{dq}{dt}=\dfrac{1}{4v^2}=\dfrac{1}{4v_0^2}$ , letting $q(0)=f(v_0)$ , we have $q=\dfrac{t}{4v_0^2}+f(v_0)=\dfrac{p}{4v^2}+f(v)$ , i.e. $v=F\left(q-\dfrac{p}{4v^2}\right)$
$v(0,q)=\pm\dfrac{1}{2\sqrt q}$ :
$F(q)=\pm\dfrac{1}{2\sqrt q}$
$\therefore v=\pm\dfrac{1}{2\sqrt{q-\dfrac{p}{4v^2}}}$
$v^2=\dfrac{1}{4\left(q-\dfrac{p}{4v^2}\right)}$
$v^2=\dfrac{v^2}{4qv^2-p}$
$4qv^4-pv^2=v^2$
$4qv^4-(p+1)v^2=0$
$(4qv^2-(p+1))v^2=0$
$4qv^2-(p+1)=0$ or $v^2=0$ (reject)
$v^2=\dfrac{p+1}{4q}$
$v=\pm\dfrac{\sqrt{p+1}}{2\sqrt q}$
$u_q=\pm\dfrac{\sqrt{p+1}}{2\sqrt q}$
$u(p,q)=\pm\sqrt{(p+1)q}+g(p)$
$u_p=\pm\dfrac{\sqrt q}{2\sqrt{p+1}}+g_p(p)$
$\therefore\left(\pm\dfrac{\sqrt q}{2\sqrt{p+1}}+g_p(p)\right)\left(\pm\dfrac{\sqrt{p+1}}{2\sqrt q}\right)=\dfrac{1}{4}$
$\dfrac{1}{4}\pm\dfrac{\sqrt{p+1}}{2\sqrt q}g_p(p)=\dfrac{1}{4}$
$g_p(p)=0$
$g(p)=C$
$\therefore u(p,q)=\pm\sqrt{(p+1)q}+C$
$u(0,q)=\pm\sqrt q+1$ :
$C=1$
$\therefore u(p,q)=\pm\sqrt{(p+1)q}+1$
$u(x,y)=y\sqrt{x^2+1}+1$
