A simple linear first order ODE 
Solve $$y'+a(t)y=b(t)$$

I try to guess my solution to be $$e^{-t\int a(s) ds}+c(t)$$
But I'm unable to solve for $c(t)$. Thank you!
 A: The solution to the equation
$y'(t) + a(t)y(t) = b(t) \tag 1$
with 
$y(t_0) = y_0 \tag 2$
is well-known to be
$y(t) = \exp \left(\displaystyle -\int_{t_0}^t a(s)ds \right) \left(y_0 + \displaystyle \int_{t_0}^t \exp \left (\displaystyle \int_{t_0}^u a(s)ds \right )b(u) du \right) \tag 3$
and this may be derived as follows:
we multiply (1) through by the integrating factor
$\displaystyle \exp \left (\int_{t_0}^t a(s) ds \right) \tag 4$
and so obtain
$y'(t)\displaystyle \exp \left(\int_{t_0}^t a(s) ds \right) + a(t) y(t)\displaystyle \exp \left(\int_{t_0}^t a(s) ds \right) = b(t)\displaystyle \exp \left(\int_{t_0}^t a(s) ds \right); \tag 5$
we note that
$\left (y(t)\displaystyle \exp \left( \int_{t_0}^t a(s) ds \right) \right )'=
\displaystyle y'(t) \exp \left( \int_{t_0}^t a(s) ds \right) + y(t)a(t) \exp \left(\int_{t_0}^t a(s) ds \right); \tag 6$
thus (5) may be written
$\left (y(t)\displaystyle \exp \left( \int_{t_0}^t a(s) ds \right) \right )' = b(t)\displaystyle \exp \left(\int_{t_0}^t a(s) ds \right); \tag 7$
we integrate both sides from $t_0$ to $t$:
$y(t)\displaystyle \exp \left( \int_{t_0}^t a(s) ds \right) - y_0 = y(t)\displaystyle \exp \left( \int_{t_0}^t a(s) ds \right) - y(t_0)\displaystyle \exp \left( \int_{t_0}^{t_0} a(s) ds \right )$
$= \displaystyle \int_{t_0}^t \left (y(u)\displaystyle \exp \left( \int_{t_0}^u a(s) ds \right) \right )'du = \displaystyle \int_{t_0}^t \exp \left (\displaystyle \int_{t_0}^u a(s)ds \right )b(u) du, \tag 8$
or
$y(t)\displaystyle \exp \left( \int_{t_0}^t a(s) ds \right) = y_0 + \displaystyle \int_{t_0}^t \exp \left (\displaystyle \int_{t_0}^u a(s)ds \right )b(u) du; \tag 9$
mutliplying (9) through by 
$\displaystyle \exp \left (-\int_{t_0}^t a(s) ds \right) \tag {10}$
yields (3).
