# Solving a PDE once you have it in canonical form.

I am working on a problem that asks me to bring the PDE $$u_{xx}+4u_{xy}+u_x=0$$ to canonical form and find the general solution.

I was able to bring the PDE to canonical form, for which I got $$u_{\zeta n}=-\frac{1}{4}u_n$$ where $$\zeta=y$$ and $$n=y-4x$$.

Now, I am not sure how to go about finding the general solution from here.

$$u_{xx}+4u_{xy}+u_x=0 \tag 1$$ Let $$\quad v=u_x$$ $$v_x+4v_y=-v \tag 2$$
$$\frac{dx}{1}=\frac{dy}{4}=\frac{dv}{-v}$$ A first characteristic equation comes from solving $$\quad \frac{dx}{1}=\frac{dy}{4}$$ : $$x-\frac{y}{4}=c_1$$ A second characteristic equation comes from solving $$\quad\frac{dy}{4}=\frac{dv}{-v}$$ : $$v e^{y/4}=c_2$$ The general solution of the PDE $$(2)$$ expressed on the form of implicit equation $$c_2=F(c_1)$$ is : $$ve^{y/4}=F\left(x-\frac{y}{4}\right)$$ $$F$$ is an arbitray function. $$u_x=e^{-y/4}F\left(x-\frac{y}{4}\right)$$ Since $$F$$ is an arbitrary function $$\quad F\left(x-\frac{y}{4}\right)=\frac{\partial}{\partial x}\Phi\left(x-\frac{y}{4}\right)\quad$$ where $$\quad\Phi$$ is an arbitrary function.
Integration wrt $$x$$ gives the general solution of PDE $$(1)$$ : $$\boxed{u(x,y)=e^{-y/4}\Phi\left(x-\frac{y}{4}\right)+\Psi(y)}$$ $$\Phi$$ and $$\Psi$$ are both arbitrary functions.
We can treat the equation firstly as a $$1$$st order PDE in $$u_n$$ to get
\begin{align*} \frac{\partial u_n}{\partial\zeta} &= -\frac{u_n}{4} \\ \implies \ln u_n &= -\frac{\zeta}{4} + \ln f(n) \\ \implies u_n &= e^{-\frac{\zeta}{4}}f(n) \\ \implies u &= e^{-\frac{\zeta}{4}}\int f(n)dn + C \end{align*}