Solving a PDE with a change of variables I have the following PDE:
$$\frac{\partial u}{\partial x}-3 \frac{\partial u}{\partial y}=x+y$$
I want to solve the equation with a change of variables; I chose the following variables:
$$\alpha:=3 x+y, \quad \beta:=x$$
So from the chain rule:
$$\begin{aligned}
&u_{x}=u_{\alpha} \alpha_{x}+u_{\beta} \beta_{x}=3 u_{\alpha}+u_{\beta}\\
&u_{y}=u_{\alpha} \alpha_{y}+u_{\beta} \beta_{y}=u_{\alpha}
\end{aligned}$$
The original PDE then becomes:
$$\begin{aligned}
&3 u_{\alpha}+u_{\beta}-3 u_{\alpha}=\alpha-2 \beta\\
&\Rightarrow u_{\beta}=\alpha-2 \beta
\end{aligned}$$
With the solution:
$$u(\alpha, \beta)=\alpha \beta-\beta^{2}+C$$
Where C is an arbitrary function.
Changing back to the original coordinates I obtain the equation:
$$u(x, y)=y x+2 x^{2}+C$$
However, I solved the same equation using methods of characteristics, and obtained the solution:
$$u(x, y)=\frac{x^{2}}{2}-\frac{y^{2}}{6}+C$$
I am sure that this last equation from the methods of characteristics is correct, because it coincides with the result given by my professor, but I do not understand why I do not get the same result using the change of variables. I have went through the equations again and again, and cannot find the mistake. Can someone help?
 A: The problem is that once you integrate the equation $u_\beta=\alpha-2\beta$ the constant is an arbitrary function in $\alpha$. Hence, $$
\qquad\qquad u(\alpha,\beta)=\alpha\beta-\beta^2+C(\alpha). \qquad \qquad (*)
$$
Then you are missing some boundary conditions to uniquely solve the PDE (in other words, to uniquely determine $C(\alpha)$). I guess with these boundary conditions you can recover your professor's solution (your procedure looks good to me, except for the integration step).
Edit: To be more clear, in case you still don't see that the constant $C$ also depends on $\alpha$: take the derivative of $(*)$ in $\beta$, you should obtain $$
\partial_\beta u=\partial_\beta\big(\alpha\beta-\beta^2+C(\alpha)\big)=\alpha-2\beta.
$$
Edit2: To support a little more my answer, notice that the function you've already found is in fact a solution. Hence, you can easily see that your professor's solution is not unique unless you have boundary conditions. This will uniquely solve the PDE.
A: When you integrated $_=−2$ with respect to  you get $(,)=−^2+$ as you've said where C is a function of  (since it must be beta independent), i.e C=f()=f(3x+y) from the change of variables. 
So your final solution is $(,)=+2^2+=yx+2x^2+f(3x+y)$, where f is an arbitrary function which satisfies the problem. So we can compare this with your other solution. 
Choosing f(3x+y)= $\frac{(3x+y)^2}{-6}$+g(3x+y) and substituting into $(,)=+2^2+=yx+2x^2+f(3x+y)$ we obtain precisely your other solution. In this case we do not have initial/boundary conditions so we cannot determine a "specific" solution to the problem. 
For example solving dy/dx=1 we get y=x+c where c is a constant, which gives an infinite family of lines. However if we require that y(0)=0 then the only solution is y=x.
