Problem with $axu_x+byu_y=P(x,y,u)$ type of quasi-linear PDE

Question

I have problem with solving quasi-linear PDE of the form $$axu_x+byu_y=P(x,y,u)$$ where a,b are constans and P is polynomial of $x,y,u.$ I start with writing Lagrange-Charpit equations $$\frac{dx}{ax}=\frac{dy}{by}=\frac{du}{P(x,y,u)}$$ and get one first integral which is $\frac{y^a}{x^b}=C_1.$ Unfortunetely I can't find second one. I do not seek for general formula, I just want to be able to solve equations like $$xu_x+2yu_y=u+x^2y\hspace{5pt}\text{or}\hspace{5pt}xu_x-yu_y=u^2(x-3y).$$

Answer 1

For $xu_x+2yu_y=u+x^2y$ ,

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$

$\dfrac{dy}{dt}=2y$ , letting $y(0)=y_0$ , we have $y=y_0e^{2t}=y_0x^2$

$\dfrac{du}{dt}=u+x^2y=u+y_0e^{4t}$ , we have $u(x,y)=\dfrac{y_0e^{4t}}{3}+e^tf(y_0)=\dfrac{x^2y}{3}+e^tf\left(\dfrac{y}{x^2}\right)$

For $xu_x-yu_y=u^2(x-3y)$ ,

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$

$\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=\dfrac{y_0}{x}$

$\dfrac{du}{dt}=u^2(x-3y)=u^2(e^t-3y_0e^{-t})$ , we have $u(x,y)=-\dfrac{1}{e^t+3y_0e^{-t}+f(y_0)}=-\dfrac{1}{x+3y+f(xy)}$