Solving Quasi Linear PDE

Question

$(x+F)\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}=F$

I dont really know what to do with this, just been stuck on it. Any help would be very much appreciated.

Answer 1

$$(x+F)\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}=F$$ System of characteristic ODEs: $\quad\frac{dx}{x+F}=\frac{dy}{1}=\frac{dF}{F}$

First family of characteristic curves , from $\quad\frac{dx}{x+F}=\frac{dF}{F}\quad\to\quad dF=\frac{F}{F+x}dx$

$F=xf(x)\quad\to\quad dF=xdf+fdx=\frac{xf}{xf+x}dx\quad\to\quad xdf=\left(\frac{f}{f+1}-f\right)dx\quad\to\quad$

$xdf=-\frac{f^2}{f+1}dx\quad\to\quad -\frac{f+1}{f^2}df=\frac{dx}{x}\quad\to\quad -\ln|f|+\frac{1}{f}=\ln|x|+c_1$

$-\ln|\frac{F}{x}|+\frac{x}{F}=\ln|x|+c_1 \quad\to\quad -\ln|F|+\frac{x}{F}=c_1$

Second family of characteristic curves , from $\quad\frac{dy}{1}=\frac{dF}{F} \quad\to\quad e^{-y}F=c_2$

General solution of the PDE expressed on the form of implicit equation : $$ \frac{x}{F}-\ln|F|=G(e^{-y}F)$$ $G$ is any differentiable function.

Answer 2

By the Lagrange method. The PDE $(x+F)F_x+F_y=F$ is converted an ODE system of equations via Lagrange's equations: $\frac{dx}{x+F}=\frac{dy}{1}=\frac{dF}{F}$. The first and third implies $\frac{dx}{x+F}=\frac{dF}{F}$.This is is just a firts order homogeneous ODE: $\frac{dx}{dF}=\frac{x+F}{F}=\frac{x}{F}+1$. Let $u=\frac{x}{F}$ $\Rightarrow \frac{dx}{dF}=u+\frac{du}{dF}$\Rightarrow $\frac{du}{dF}=1$ $\Rightarrow u=F+c_1$ $\Rightarrow \frac{x}{F}=F+c_1$ or $c_1=\frac{x}{F}-F$ (one of the first integrals). For the second first integral we take the second and third $\frac{dy}{1}=\frac{dF}{F}$ which is a separable ODE with the solution $\ln|F|=c_2+y$ or $c_2=\ln|F|+y$. Now the general solution is expressed implicityl as $c_1=g(c_2)$ or

$\frac{x}{F}-F=g(\ln|F|+y)$, where $g$ is an arbitrary $C^1$-function.