Understanding a non-autonomous ODE Consider $x'(t) = a(t)x$
a) Find a formula involving integrals for the solution of this system.
b) Prove that your formula gives the general solution of this system.
I am new to ODE's and we have gone over integrating the system...
$x'/x = a(t)$ in order to attempt to isolate x in terms of $t$.  I thus get $log(x) = 0.5at^2+C \; \text{(a constant)}$, which yields $x(t) = e^(.5at^2+C)$, not sure if I'm right here.  I'm also not sure why it says "involving integrals," making it sound like the solution needs integral signs in it?
For part b, I have no idea how to even start it, not sure what I need to show.  
Any help appreciated!
 A: From
$x'(t) = a(t)x(t) \tag 1$
we write
$(\ln x(t))' = \dfrac{x'(t)}{x(t)} = a(t); \tag 2$
whence
$\ln \left ( \dfrac{x(t)}{x(t_0)} \right ) = \ln x(t) - \ln x(t_0) = \displaystyle \int_{t_0}^t \dfrac{x'(t)}{x(t)} \; dt = \int_{t_0}^t a(t) \; dt, \tag 3$
or
$\dfrac{x(t)}{x(t_0)} = \exp \left ( \displaystyle \int_{t_0}^t a(t) \; dt \right ), \tag 4$
or
$x(t) = x(t_0) \exp \left ( \displaystyle \int_{t_0}^t a(t) \; dt \right ), \tag 5$
which solves part (a).
We note that the division by $x(t)$ occurring in (2) is legitimate when $x(t) \ne 0$.  If $x(t_0) = 0$ for some $t_0$, then $x(t) = 0$ is the unique solution to (1) with $x(t_0) = 0$.  Therefore a solution initialized with $x(t_0) > 0$ will remain positive for all $t$; likewise a solution with $x(t_0) < 0$ will remain negative, and thus step (2) is valid, as is corroborated by equation (5).
For part (b) We have shown that $x(t)$ satisfying (1) with $x(t_0) \ne 0$ at some $t$ must be of the form (5); (5) is a necessary condition for (1) to apply.  Therefore, (5), being the only solution, is completely general in the sense that it binds for any initial condition $x(t_0)$; it is as general as one could wish.  
A: If you have $$x'(t) = a(t)\,x(t)$$ then
$$\frac{x'(t)}{x(t)}=a(t)$$
$$\int\frac{x'(t)}{x(t)}\,dt=\int a(t)\,dt \implies \log (|x(t)|)=\int a(t)\,dt+C\implies x(t)=C \, e^{\int a(t)\,dt }$$
