Let this differential equation :
$$(H)\qquad x''+tx'-(1+t)x=0$$
As $\phi_a=e^t$, is a solution of $(H)$, we can deduce $\phi_b$ from the wronskian $W(t)=\begin{array}{|ll|}\phi_a(t)&\phi_b(t) \\ \phi_a'(t) & \phi_b'(t)\end{array}$ Thus $\left(\dfrac{\phi_b(t)}{\phi_a(t)}\right)'=\dfrac{W(t)}{(\phi_a(t))^2}$
As $W(t)$ is a solution of this equation $y'(t)=-t\cdot y(t)$ we have got $\displaystyle W(t)=\exp\bigg[-\int_{t_0}^t s\cdot ds\bigg]=e^{-t^2/2+t_0^2/2}=\lambda e^{-t^2/2}\qquad \lambda :=e^{t_0^2/2}$
So $\left(\dfrac{\phi_b(t)}{\phi_a(t)}\right)'=\dfrac{\lambda e^{-t^2/2}}{e^{2t}}=\lambda\cdot e^{-t^2/2-2t}$
And now I'am blocked, because if $\displaystyle \gamma(t)=\int\lambda\cdot e^{-t^2/2-2t}\cdot dt$, we have got $\dfrac{\phi_b(t)}{\phi_a(t)}=\gamma(t)$
So $\phi_b(t)=\phi_a(t)\cdot \gamma(t)$, how to calculate $\gamma(t)$???
Since $x_1=e^t$ is a solution than try $x=x_2e^t$
then it reduces to $$x_2''+2x_2'+tx_2'=0$$ Substitute $u=x_2'$ $$u'+u(t+2)=0 \implies u=Ke^{-\frac {t^2}2-2t}$$ unfortunately it has a Gaussian like integral in it : $$x_2(t)=K_1\int e^{-\frac {t^2}2-2t}+K_2$$ $$\boxed{x(t)=K_1e^t\int e^{-\frac {t^2}2-2t}+K_2e^t}$$