Answer
$y=\dfrac{3}{2}-\dfrac{1}{2}e^{-2t}$
Work Step by Step
Here, we have $y'+2y=3$
The integrating factor is: $I=e^{\int 2 dt}=e^{2t}$
Now, $e^{2t}[y'+2y]=3e^{2t}$
and $\int [ye^{2t}]' dt=\int 3 e^{2t}$
$\implies ye^{2t}=\dfrac{3}{2}e^{2t}+c$
Next,apply the initial conditions in the above equation, we have
$y[e^{0}]=\dfrac{3e^{0}}{2}+c$
or, $c=-(\dfrac{1}{2})$ and $ye^{2t}=\dfrac{3e^{2t}}{2}-\dfrac{1}{2}$
so, the general solution is:
$y=\dfrac{3}{2}-\dfrac{1}{2}e^{-2t}$
