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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!

From

$x'(t) = a(t)x(t) tag 1$

we write

$(ln x(t))' = dfracx'(t)x(t) = a(t); tag 2$

whence

$ln left ( dfracx(t)x(t_0) right ) = ln x(t) - ln x(t_0) = displaystyle int_t_0^t dfracx'(t)x(t) ; dt = int_t_0^t a(t) ; dt, tag 3$

or

$dfracx(t)x(t_0) = exp left ( displaystyle int_t_0^t a(t) ; dt right ), tag 4$

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.

If you have $$x'(t) = a(t),x(t)$$ then
$$fracx'(t)x(t)=a(t)$$
$$intfracx'(t)x(t),dt=int a(t),dt implies log (|x(t)|)=int a(t),dt+Cimplies x(t)=C , e^int a(t),dt $$