Answer
$y\rightarrow\frac{3}{2}$ as $t\rightarrow\infty$.
Work Step by Step
Given $y' = 3-2y$, by means of a direction field, determine the behavior of $y$ as $t \rightarrow \infty$.
Since $y' = 3-2y$, the slope of $y$ as a function of $t$ is $0$, is determined by setting $0 = 3-2y$ and solving for $y$: I.e., $y = \frac{3}{2}$.
We can now "sketch" a direction field:
We know that at $y = \frac{3}{2}$, the line segments representing the slope of any graph representing a solution of our differential equation intersecting the horizontal line at $y = 0$ is $0$.
Also, the slope of the line segments along the horizontal line at $y =2$ can be determined from $3-2\cdot 2 = -1$.
Likewise, the slope of the line segments at the corresponding horizontal line at $y = 1$ is determined by $3-2\cdot 1 = 1$, So our direction field for these three different values for $y$ would be the following:
$2$ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
$\frac{3}{2}$ ___________________________
$1$ ////////////////////////////////////////
Slopes for the line segments further away (either positive or negative) from the horizontal line $y = \frac{3}{2}$ would be steeper as we would know frem putting in ( smaller or larger) values for $y$ than $2$ or $1$, respectively.
By inspecting our direction field, we see that $y\rightarrow\frac{3}{2}$ as $t\rightarrow\infty$. Also, as noted, we see that the further away that $y$ is from $\frac{3}{2}$, the more rapidly our solution to the given differential equation will approach the asymptote $y =\frac{3}{2}$.
