General solution of a PDE- Lagrange or Characteristics method I am trying to find the general solution of the PDE:
$$xu_x + (1+y)u_y= x(1+y)+xu$$


*

*If the initial condition is $$u(x,6x-1)=\phi(x)$$
then what is the necessary condition for $\phi$ that guarantees the existence of a solution? How can one solve the problem for the appropriate functions $\phi$.


*The same question as 1 if we change the initial condition to $$u(-1,y)=\psi(y)$$
What are the differences between 1 and 2?
I apologize in advance for asking in this way , because I am totally new to this subject and I am trying to learn some ideas.
Thanks an advance.
 A: $$xu_x+(1+y)u_y=x(1+y)+xu$$
Charpit-Lagrange equations :
$$\frac{dx}{x}=\frac{dy}{1+y}=\frac{du}{x(1+y)+xu}=ds$$
A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{1+y}$ :
$$\frac{1+y}{x}=c_1$$
A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{x(1+y)+xu}=\frac{du}{x(c_1x)+xu}$
This is a first order linear ODE : $\frac{du}{dx}-u=c_1x$
$$ue^{-x}+c_1(x+1)e^{-x}=c_2$$
$c_1$ and $c_2$ are arbitrary related which leads to the general solution on the form of implicit equation :
$$ue^{-x}+\frac{1+y}{x}(x+1)e^{-x}=F\left(\frac{1+y}{x}\right)$$
$F$ is an arbitrary function (to be determined according to boundary conditions).
$$u(x,y)=-\frac{1+y}{x}(x+1)+e^xF\left(\frac{1+y}{x}\right)$$
FIRST CASE of boundary condition :
$u(x,6x-1)=\phi(x)=-\frac{1+(6x-1)}{x}(x+1)+e^xF\left(\frac{1+(6x-1)}{x}\right)$
$\phi(x)=-6(x+1)+e^xF\left(6\right)$
This implies that the function $\phi(x)$ must have a particular form :
$$\phi(x)=-6(x+1)+e^xC\quad\text{where}\quad C=\text{constant}$$
In general the functions $\phi(x)$ have not this very particular form and as a consequence the function $F$ cannot be determined. The problem has no solution fitting to the boundary condition. If by luck the function $\phi(x)$ has the above particular form, the solution is $\quad u(x,y)=-\frac{1+y}{x}(x+1)+e^xC$ 
SECONDCASE of boundary condition :
$u(-1,y)=\psi(y)=-\frac{1+y}{-1}(-1+1)+e^{-1}F\left(\frac{1+y}{-1}\right)=e^{-1}F(-1-y)$
Let $X=-1-y\quad;\quad y=-X-1$
$\psi(-X-1)=e^{-1}F(X)$
The function $F(X)$ is determined :
$$F(X)=e\:\psi(-X-1)$$
Now we can put it into the general solution where $X=\frac{1+y}{x}$ :
$$u(x,y)=-\frac{1+y}{x}(x+1)+e^x e\:\psi\left(-\frac{1+y}{x}-1\right)$$
In this case, the problem has a well determined solution fitting to the boundary condition :
$$u(x,y)=-\frac{1+y}{x}(x+1)+e^{x+1}\psi\left(-\frac{1+y+x}{x}\right)$$
A: The idea of the method of characteristics is to find curves on the plane (characteristics) where solutions to the original PDEs satisfy a simple ODE. Let's see how thiw works here :
Let $\gamma=(x(s);y(s))$ be a parametrised curve. Then, along $\gamma$ any solution the original PDE satisfies
$$\begin{array}{}
xu_x+(1+y)u_y&=&x(1+y)+xu\\
x'u_x+y'u_y&=&du
\end{array}$$
Therefore if $x'=x$ and $y'=1+y$ then $u$ satisfies a nice ODE along $\gamma$, namely : $$\frac{du}{ds}=x(s)(1+y(s))+x(s)u(s).$$
Now, it's easy to see that such $\gamma$ are just half-lines starting from $(0,-1)$ (draw them !).
But $y=6x-1$ is a line going through $(0,-1)$! 
So, $u$ and therefore $\phi$ must satisfy the following ODE (where $s\in\mathbb{R}$):
$$\begin{array}{}
\phi'(s)&=&s(6s-1+1)+s\phi(s)\\
&=&6s^2+s\phi(s)
\end{array}
$$
The point is that $y=6x-1$ isn't transverse to the characteristic curves it meets. So there are further constraints on the kind of initial conditions we can consider.
2) By contrast, the line $x=-1$ is transverse to every characteristic curves it meets (draw it !), so there's no constraint on $\psi$.
