#### Answer

2. Statement of Problem: Given $y'=2y-3$, a) draw the associated direction field; b) determine the behavior of $y$ as $t\rightarrow\infty$; c) if this behavior depends on the initial value of $y$ at $t=0$, describe the dependency.
Answer:
a)
$2$
$\frac{3}{2}$
$1$
$0$
b)
$y\rightarrow\frac{+}{-}\infty$ as $t\rightarrow\infty$.
c) The further away (in either direction) that $y(0)$ is from $\frac{3}{2}$ the faster the function $y(t)$ diverges (to $-\infty$ or $+\infty$). Specifically, if $y(0)\frac{3}{2}$, $y\rightarrow +\infty$.

#### Work Step by Step

Solution to 2. a): Since we are asked to "draw" the direction field and only have a standard keyboard, it suffices here to represent the line segments at a given $y$ value (on the vertical axis) by a slash $(/)$ if they are to have a positive slope, a backslash ( \ ) if they have a negative slope and an underscore ( _ ) for the line segments having $0$ slope. (We imagine the line segments having steeper slopes the further away from the line at $y=\frac{3}{2}$ they are.)
This is how we obtain the values for the slopes of the line segments in our direction field for the given equation, $y'=2y-3$: We note that $y'$, denoting the derivative of the function $y$, has for a geometric interpretation, the instantaneous slope of the graph for $y(t)$. Thus, when the slope of the graph for $y$ is to be $0$, we have from our given differential equation, $0=2y-3$. Solving for $y$ we get $y=\frac{3}{2}$. So when our graph has $y$ value $\frac{3}{2}$, it will have an instantaneous slope of $0$.
The slopes other than $0$ for various values of $y$ can be found immediately, again, from our differential equation. Thus,
$2(3)-3=3$,
$2(2)-3=1$,
$2(1)-3=-1$,
$2(0)13=-3$.
b) By inspecting our direction field for this equation we may conclude that if our graph intersects the $y$-axis above $\frac{3}{2}$, then $y\rightarrow+\infty$ as $t\rightarrow\infty$. And if our graph intersects the $y$-axis below $\frac{3}{2}$, then $y\rightarrow - \infty$ as $t\rightarrow\infty$.
c) Also, by inspecting our direction field (including the imagined line segments with steeper slopes for $y$ values further away from $\frac{3}{2}$), we can conclude that the further away that $y(0)$ is from $\frac{3}{2}$ the faster that $y(t)$ diverges to (+or-) $\infty$.