Solving $(y+u) u_x + y u_y = x - y $

Question

I have the following problem:

$$(y+u) u_x + y u_y = x - y $$

I only know characteristic curves so I did:

$\frac{dx}{dt} = y + u $

$\frac{dy}{dt} = y$

I began solving the second equation resulting $y = C_1 e^t$. Using that result I solved $\frac{dx}{dt} = C_1 e^t + u $. This is where my doubts began, since $x$ has $u$. I decided to give it a try so I got $x = C_1 e^t + u t $. Here there's a constant that I've been said the method omits. And I've seen some examples where it's omitted. So I continued by using the chain rule, ending up in this equation:

$$\frac{du}{dt} = x - y = (C_1 e^t + u t) - C_1 e^t = u t $$

Solving this gives $u = e^{\frac{t^2}{2}} + K $ where I know $K$ is an arbitrary function. Replacing $t$ from $y = C_1 e^t$ we get $$u = \ln \big(\frac{y}{C_1}\big) + K $$ but clearly this is incomplete. I don't know the form of $K(x,y)$ and $C_1$. I would like to know how to manage the arbitrary functions since I feel it's what's happening, and how to manage them in this case where we have a quasilinear PDE. Also, if there's another way to solve this, I don't know if substracting $\frac{dx}{dt}$ and $\frac{dy}{dt}$ helps in some way since it leaves $u$ or if there's a common change of variables that's useful in these problems. Thanks for the help.

Answer 1

$$(y+u) u_x + y u_y = x - y $$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{y+u}=\frac{dy}{y}=\frac{du}{x-y}$$

FIRST :

$\frac{dx}{y+u}=\frac{du}{x-y}=\frac{dx+du}{(y+u)+(x-y)}=\frac{dx+du}{x+u}$

A first characteristic equation comes from solving $\quad \frac{dy}{y}=\frac{dx+du}{x+u}$ $$\frac{u+x}{y}=c_1$$

SECOND :

$\frac{dy}{y}=\frac{du}{x-y}=\frac{dy+du}{(y)+(x-y)}=\frac{dy+du}{x}$

A second characteristic equation comes from solving $\quad \frac{dx}{y+u}=\frac{dy+du}{x}$ $$(y+u)^2-x^2=c_2$$ GENERAL SOLUTION :

The general solution of the PDE can be expessed on the form of various equivalent implicit equations , for example : $$\Phi\left(\frac{u+x}{y}\,,\,\left((y+u)^2-x^2\right)\right)=0$$ where $\Phi$ is an arbitrary function of two variables.

Or : $$\frac{u+x}{y}=F\big((y+u)^2-x^2\big)$$ where $F$ is an arbitrary function.

Or : $$(y+u)^2-x^2=G\big(\frac{u+x}{y}\big)$$ where $G$ is an arbitrary function.

Other equivalent implicit equations can be presented with other arbitrary functions involving $c_1$ and $c_2$ , all related one to another.

Some boundary condition has to be specified to determine one of those functions and thus the solution satisfying both the PDE and the condition.

NOTE :

If you prefer the method with parameter you can write $$\frac{dx}{y+u}=\frac{dy}{y}=\frac{du}{x-y}=dt$$ Separate the three ODEs and combine them in order to solve for functions of the parameter $t$.