## University Calculus: Early Transcendentals (3rd Edition)

$y=\dfrac{3}{2}-\dfrac{e^{-2t}}{2}$
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: $y=\dfrac{3}{2}-\dfrac{e^{-2t}}{2}$