nonhomogenious partial differential equation How to solve this transport equation?
$\dfrac{u_t}{u}-\dfrac{t}{x}\dfrac{u_x}{u}=-\dfrac{fx^2+2g}{ax^2+2b}$   in $(0,\infty)\times \mathbb{R}-\{0\}$
$u(t_0,x)=u_{t_0}(x)  $ on  $\{t=t_0\}\times \mathbb{R}-\{0\}$
where $u=u(t,x)$  is a real function of two variables $t,x$ and $a=a(t)$, $b=b(t)$, $f=f(t)$ and $g=g(t)$ are real functions of $t\in \mathbb{R}$.
Please help me.
 A: $\dfrac{u_t}{u}-\dfrac{t}{x}\dfrac{u_x}{u}=-t\dfrac{f(t)x^2+2g(t)}{a(t)x^2+b(t)}$
$u_t-\dfrac{t}{x}u_x=-\dfrac{tf(t)x^2+2tg(t)}{a(t)x^2+b(t)}u$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=1$ , we have $t=s$
$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=1$ , we have $t=s$
$\dfrac{dx}{ds}=-\dfrac{t}{x}=-\dfrac{s}{x}$ , letting $x(0)=x_0$ , we have $x^2=x_0^2-s^2=x_0^2-t^2$
$\dfrac{du}{ds}=-\dfrac{tf(t)x^2+2tg(t)}{a(t)x^2+b(t)}u=-\dfrac{sf(s)(x_0^2-s^2)+2sg(s)}{a(s)(x_0^2-s^2)+b(s)}u$ , we have $u(t,x)=F(x_0^2)e^{-\int_{s_0}^s\frac{rf(r)(x_0^2-r^2)+2rg(r)}{a(r)(x_0^2-r^2)+b(r)}dr}=F(x^2+t^2)e^{-\int_{t_0}^t\frac{rf(r)(x^2+t^2-r^2)+2rg(r)}{a(r)(x^2+t^2-r^2)+b(r)}dr}$
$u(t_0,x)=u_{t_0}(x)$ :
$F(x^2+t_0^2)=u_{t_0}(x)$
$F(x)=u_{t_0}\left(\pm\sqrt{x-t_0^2}\right)$
$\therefore u(t,x)=u_{t_0}\left(\pm\sqrt{x^2+t^2-t_0^2}\right)e^{-\int_{t_0}^t\frac{rf(r)(x^2+t^2-r^2)+2rg(r)}{a(r)(x^2+t^2-r^2)+b(r)}dr}$
