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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.

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$$u_{xx}+4u_{xy}+u_x=0 \tag 1$$ Let $\quad v=u_x$ $$v_x+4v_y=-v \tag 2$$

This is a linear first order PDE. The Charpit-lagrange system of characteristic ODEs is :

$$\frac{dx}{1}=\frac{dy}{4}=\frac{dv}{-v}$$ 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.

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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*}

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