Answer
$y(t) = 3 t^2 - 12 t + 24 + C e^{-t/2}, \quad y \to \infty \text{ as } t \to \infty$
Work Step by Step
Solve:
$2y' + y = 3t^2$
and analyze as $t \to \infty$.
Standard linear form
$y' + \frac{1}{2}y = \frac{3}{2} t^2$
$P(t) = \frac{1}{2}$
Integrating factor
$\mu(t) = e^{\int P(t) dt} = e^{t/2}$
Multiply ODE
$e^{t/2} y' + \frac{1}{2} e^{t/2} y = \frac{3}{2} t^2 e^{t/2}$
$\frac{d}{dt}(e^{t/2} y) = \frac{3}{2} t^2 e^{t/2}$
Integrate
$e^{t/2} y = \frac{3}{2} \int t^2 e^{t/2} dt$
Evaluate $\int t^2 e^{t/2} dt$
$\int t^2 e^{t/2} dt = 2 t^2 e^{t/2} - 8 t e^{t/2} + 16 e^{t/2}$
Substitute
$e^{t/2} y = \frac{3}{2}(2 t^2 e^{t/2} - 8 t e^{t/2} + 16 e^{t/2}) + C$
$e^{t/2} y = 3 t^2 e^{t/2} - 12 t e^{t/2} + 24 e^{t/2} + C$
Solve for $y$
$y = 3 t^2 - 12 t + 24 + C e^{-t/2}$
Behavior as $t \to \infty$
$C e^{-t/2} \to 0$, and $3 t^2 - 12 t + 24 \to \infty$
$\Rightarrow y \to \infty$
