Have I obtained the proper solution to this PDE? I'm a little stuck on this.
Consider $ u_t -(1+t^2)u_x = \phi(x,t) \quad  u(x,0)=u_0(x)$
Via the method of characteristics, the total derivative of $u(x,t)$ is 
$$\frac{du}{dt} = \dfrac{\partial u}{\partial t} + \frac{dx}{dt}\frac{\partial u}{\partial x}\>. $$
Therefore, the characteristics satisfy
$$\frac{dx}{dt} = -(1+t^2) \implies x(t)=-\left(t+\frac{t^3}{3}\right) +x_0\>, $$ 
And the value of $u$ satisfies
$$ \frac{du}{dt} = \phi(t)\>.$$
Here, the $x$ dependance has been dropped, as $x$ is a function of $t$.  
Now, I could write the solution to this DE as 
$$u(x(t),t) = u_0(t_0) + \int_{t_0}^{t} \phi(z) dz = u_0(t_0) -\Phi(t_0) + \Phi(t)$$
But that doesn't seem to get me anywhere.  The solution should have the initial profile moving along the characteristic, with the magnitude of the profile changing according to the solution to the differential equation for $u$ w.r.t $t$.
So how can I rectify this?

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dx}{ds}=-(1+t^2)=-1-s^2$ , letting $x(0)=x_0$ , we have $x=x_0-s-\dfrac{s^3}{3}=x_0-t-\dfrac{t^3}{3}$
$\dfrac{du}{ds}=\phi(x,t)=\phi\left(x_0-s-\dfrac{s^3}{3},s\right)$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)+\int_0^s\phi\left(x_0-r-\dfrac{r^3}{3},r\right)dr=f\left(x+t+\dfrac{t^3}{3}\right)+\int_0^t\phi\left(x+t-r+\dfrac{t^3-r^3}{3},r\right)dr$
$u(x,0)=u_0(x)$ :
$f(x)=u_0(x)$
$\therefore u(x,t)=u_0\left(x+t+\dfrac{t^3}{3}\right)+\int_0^t\phi\left(x+t-r+\dfrac{t^3-r^3}{3},r\right)dr$
