How to solve this one order quasi linear PDE?

Question

I got a PDE: $f_{y}=axf^{2}f_{x}-bff_{x}-\frac{f^{2}}{x}$ where a, b are constant . I want a solution of $f(x,y)$. and i try to covert it into linear one, but failed . could you help me?

Answer 1

$f_y=axf^2f_x-bff_x-\dfrac{f^2}{x}$

$(axf^2-bf)f_x-f_y=\dfrac{f^2}{x}$

$(af^2x^2-bfx)f_x-xf_y=f^2$

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

$\dfrac{df}{dt}=f^2$ , we have $-\dfrac{1}{f}=t$ , i.e. $f=-\dfrac{1}{t}$

$\dfrac{dx}{dt}=af^2x^2-bfx=\dfrac{ax^2}{t^2}+\dfrac{bx}{t}$

Let $x=tu$ ,

Then $\dfrac{dx}{dt}=t\dfrac{du}{dt}+u$

$\therefore t\dfrac{du}{dt}+u=au^2+bu$

$t\dfrac{du}{dt}=au^2+(b-1)u$

$\dfrac{du}{au^2+(b-1)u}=\dfrac{dt}{t}$

$\int\dfrac{du}{(au+(b-1))u}=\int\dfrac{dt}{t}$

$\dfrac{1}{b-1}\ln\dfrac{u}{au+b-1}=\ln t+c$

$\ln\dfrac{-xf}{-axf+b-1}=(b-1)\ln t+c$

$\dfrac{xf}{axf-b+1}=t^{b-1}x_0$

$\dfrac{xf}{axf-b+1}=\dfrac{(-1)^{b-1}x_0}{f^{b-1}}$

$x_0=\dfrac{(-1)^{b-1}xf^b}{axf-b+1}$

$\dfrac{dy}{dt}=-x$ , letting $y(0)=g(x_0)$ , we have $y=g(x_0)-xt=g\left(\dfrac{(-1)^{b-1}xf^b}{axf-b+1}\right)+\dfrac{x}{f}$