Answer
$y=\dfrac{3}{2}-\dfrac{e^{-2t}}{2}$
Work Step by Step
The standard form of the given equation is:
$y'+2y=3$ ....(1)
The integrating factor is: $e^{\int 2 dt}=e^{2t}$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [ye^{2t}]' dt=\int 3 e^{2t}$
or, $ye^{2t}=\dfrac{3}{2}e^{2t}+c$
Applying the initial conditions in the above equation, we get
$ye^{0}=\dfrac{3}{2}e^{0}+c \implies c=\dfrac{-1}{2}$
and $ye^{2t}=\dfrac{3}{2}e^{2t}-\dfrac{1}{2}$
Thus, the general solution is as follows:
$y=\dfrac{3}{2}-\dfrac{e^{-2t}}{2}$
