Answer
$y_0=-16/3$
Work Step by Step
Finding the integrating factor:
$\mu(t)=\exp(\int -3/2\ dt)$
$\mu(t)=e^{-3t/2}$
Finding the y function:
$y(t)=1/\mu(t) [\int \mu(s)(3s+2e^s)ds+c]$
$y(t)=e^{3t/2}[-2/3\ e^{-3t/2}(3t+6e^t+2)+c]$
$y(t)=-2t-4e^t-4/3+ce^{3t/2}$
$y(0)=-4-4/3+c=y_0\rightarrow c=y_0+16/3$
The term with the bigger impact on the limit at infinity is $ce^{3t/2}$, because it's an exponential function (grows faster than just t) with a bigger argument than $e^t$ for the same value of t, so the value of $y_0$ that divides y growing positively or negatively approaching infinity is $y_0=-16/3$
