I cannot find my error in the following exercise:
In a previous exercise I showed that the IVP $$\dot x(t)=f(t)x(t)+g(t),\ x(t_0)=x_0$$is solved by a certain $x(t)=C(t)e^{F(t)}$ where $\dot F=f$ and $F(t_0)=0$. I plugged this into the equation and got $\dot C(t)=g(t)e^{-F(t)}$ and concluded that $$C(t)=\int_{t_0}^tg(t')\exp(-F(t'))dt'+\underbrace{C(t_0)}_{=x_0}$$
Now, either this is wrong, or there is an error in my try to apply this to the IVP $$\dot x=\frac{2x}{t}+2t^3,\ x(2)=20$$ We have $t_0=2,x_0=20,f(t)=\frac{2}{t},F(t)=2\log t,g(t)=2t^3$, so plugging all of this into my $C(t)$ I obtain $$C(t)=20+\int_2^t2t'^3\exp(-2\log(t'))dt'=20+\int_2^t2t'dt'=t^2+16$$ Obviously my $x(t)=t^4+16t^2$ then solves the equation, but the starting value is totally off: $x(2)=80$
You based your constant $C(t_0)=x_0$ on the assumption that $e^{-F(t_0)}=1$, which it isn't for your chosen $F$, so you need to either modify $F$ so that $F(2)=0$, or correct the value of the constant of integration in $C$.