Answer
a) The direction field has an equilibrium solution of $y=-(\frac{1}{2})$. Some solutions for $y′=-1-2y$ are: $y′(0)=-1$, $y′(1)=-3$, $y′(-1)=1$, $y'(-\frac{1}{2})=0$, and $y′(-2)=3$. The directional field should be convergent for increasing $t$
b) As $t →∞$, $y→-\frac{1}{2}$ for all $y_0$.
c) The behavior of $y$ as $t →∞$ does not depend upon initial conditions.
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 $t$. Positive values of $y'$ indicate that $y$ is increasing as t increases (at that specific point), negative indicates the opposite.
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 $y′=a+by$, let $y′=0$, then solve the right side of the original equation for $y$. This gives the equilibrium solution $y_{eq}=-\frac{a}{b}$. In a graph of $y$ vs. $t$, this will appear as a horizontal line; 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: "Some solutions for $y′=-1-2y$ are: $y′(0)=-1$, $y′(1)=-3$, $y′(-1)=1$, $y'(-\frac{1}{2})=0$, and $y′(-2)=3$." Graphing all of the values here, along with the equilibrium solution $y_{eq}$, 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) As $t →∞$, the behavior of $y$ depends upon the initial conditions (the value of $y_0$). For the given equation however, as $t →∞$, $y→-\frac{1}{2}$ for all $y_0$, so the end behavior is not dependent on initial conditions. (further explanation in part c)
c)There are three possible behaviors of $y$:
If $y_0=-\frac{1}{2}$, then the value of y does not change because $y′(-\frac{1}{2})=0$; $y=-\frac{1}{2}$ for all $t$. As $t →∞$, $y→-\frac{1}{2}$.
If $y_0\gt-\frac{1}{2}$, then as $t →∞$, $y→-\frac{1}{2}$. Consider some values of $y′$ if $y\gt-\frac{1}{2}$: $y′(1)=-3$, $y′(0)=-1$. In all cases where $y\gt-\frac{1}{2}$, $y′\lt0$. Notice that for values that are closer to $-\frac{1}{2}$, the magnitude of $y′$ is smaller. If graphs are traced based on the values of $y′$, you will find that no equation $y$ will ever cross the line $y_{eq}$. The same is true for all cases where $y_0\lt-\frac{1}{2}$: as $t →∞$, $y→-\frac{1}{2}$. Consider some values of $y′$ if $y_0\lt-\frac{1}{2}$: $y′(-1)=1$, $y′(-2)=3$. These values are positive, indicating that $y$ is increasing as $t$ is increasing; $y$ is approaching $-\frac{1}{2}$. The magnitude of $y′$ decreases as $y→-\frac{1}{2}$. This is because as $t →∞$, $y→-\frac{1}{2}$.
