Answer
a) The direction field has equilibrium solutions of $y=0$ and $y=5$. Some solutions for $y′=-y(5-y)$ are: $y′(0)=0$, $y′(4)=-4$, $y′(5)=0$, $y'(6)=6$, and $y′(-1)=6$ .
b) As $t →∞$, the behavior of y depends upon the initial conditions (the value of $y_0$).
c) There are 5 distinct starting states of $y$, with 3 distinct behaviors as $t →∞$:
If $5\gt{y_0}$, then as $t →∞$, $y→0$.
If $y_0=5$, then as $t →∞$, $y→5$.
If $y_0\gt5$, then as As $t →∞$, $y→∞$
Work Step by Step
Note: $y'$ is a function of $y$. In essence, $y′=y′(y)$. Additionally, $y=y(t)$. Directional fields of differential equations give a value for $\frac{dy}{dt}$ for a set of values of $y$ and $t$.
a) Start graphing a directional field by finding any equilibrium solutions. To find the equilibrium solution (called $y_{eq}$) of a first order differential equation $y′=a+by+cy^2$, let $y′=0$, then solve the right side of the original equation for $y$. This gives the equilibrium solutions $y_{eq1}=\frac{-b-\sqrt {b^2-4ac}}{2a}$ and $y_{eq2}=\frac{-b+\sqrt {b^2-4ac}}{2a}$. In a graph of $y$ vs. $t$, these will appear as horizontal lines; any equation $y(t)$ that "touches" this line for a given value of $t$ will continue along the line for all greater values of $t$. Since $y'$ does not depend on $t$, $y'$ will be the same at a given value of $y$ for all values of t. Find some values of $y'$ for values of $y$. Some sample answers are given above in the answer to part a). Graphing all of the values here, along with the equilibrium solutions, $y_{eq1}$ and $y_{eq2}$, should give the directional field enough information to see the general behavior of $y$. REMEMBER that for the given equation, all values of $t$ for a given $y$ have the same value for $y′$.
b) Pick several starting points: some below, some above, some between, and some along $y_{eq1}$ and $y_{eq2}$. Trace an approximate graph of a particular $y$ in each aforementioned region. Based on the values of $y'$that were calculated, observe the behavior of each particular $y$ as $t →∞$. The behavior of y should be apparent: it will either approach $∞$, $-∞$, or it will approach a value equal to one of the equilibrium values ($y=0$ and $y=5$ in this case).
c) There are 5 distinct starting states of $y$, with 3 distinct behaviors as $t →∞$:
If $y_0\gt5$, then as $t →∞$, $y→∞$. This is because all values of $y'$ are positive and increase in magnitude with increasing $y$.
If $y_0=5$, then the value of y does not change because $y'=0$ for all $t$; $y=5$ for all $t$. As $t →∞$, $y→5$.
If $5\gt{y_0}\gt0$, then as $t →∞$, $y→0$. This is because within this range, $y'$ is negative , so any particular $y$ will always approach 0.
If $y_0=0$, then the value of $y$ does not change. The value of y does not change because $y'=0$ for all t. As $t →∞$, $y→0$.
If $y_0\lt0$, then as As $t →∞$, $y→0$. This is because $y'$ is positive for all $y_0\lt0$. Any particular $y$ will only increase as $t$ increases.
