Don't understand the last part of this question:
Find the value of $y_0$ for which the solution of the initial value problem:
$$y'-y=1+3\sin t , \ \ y(0)=y_0$$
remains finite as $t\to \infty$.
Attempt: I know I have to find the general solution so using an integrating factor ($e^-t$), I figured out the general solution is:
$$y(t) = ce^t -\frac32\sin t-\frac32 \cos t-1.$$
then plugging $0$ for $t$ gives:
$$y(0) = -\frac52 + ce^t.$$
I don't know how to find the value of $y_0$ or what it really means. I understand $c$ must be 0 as $t \to \infty$ or else it will be unbounded but don't know how to follow through.
Since you do not know $y_0$, the condition $y(0) = y_0$ only give
$$ y_0 = -\frac 52 + c,$$
or $c = \frac 52 + y_0$. So the solution is
$$y(t) = \left( \frac 52 + y_0\right)e^t -\frac32\sin t-\frac32 \cos t-1.$$
Now you want to know if $y(t)$ remains finite as $t \to \infty$. The term
$$-\frac32\sin t-\frac32 \cos t-1$$
is bounded for all $t$, so the only problem occurs at the $e^t$ terms. If by $t\to \infty$ you mean positive infinity, then
$$\left( \frac 52 + y_0\right)e^t$$
is not bounded unless the coefficient is zero. That is, $y_0 = -\frac 52$.