Answer
$y = 3t - 6 + Ce^{-t/2}$
Work Step by Step
$$2y' + y = 3t$$
Divide both sides by 2:
$$y' + \tfrac{1}{2}y = \tfrac{3}{2}t$$
Integrating factor:
$$\mu(t) = e^{\int P(t),dt} = e^{\int \tfrac{1}{2},dt} = e^{t/2}$$
Multiply through by $\mu(t)$:
$$e^{t/2}(y' + \tfrac{1}{2}y) = e^{t/2}\tfrac{3}{2}t$$
$$(e^{t/2}y)' = \tfrac{3}{2}t e^{t/2}$$
Integrate both sides:
$$\int (e^{t/2}y)',dt = \tfrac{3}{2}\int t e^{t/2},dt$$
Integration by parts:
$$\int t e^{t/2},dt = 2t e^{t/2} - 4e^{t/2} + C$$
Therefore:
$$e^{t/2}y = 3t e^{t/2} - 6e^{t/2} + C$$
Divide by (e^(t/2)):
$$y = 3t - 6 + Ce^{-t/2}$$
