Chapter 1 - Introduction - 1.1 Some Basic Mathematical Models; Direction Fields - Problems - Page 7: 5

a) The direction field has an equilibrium solution of $y_{eq}=-(\frac{1}{2})$. 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 field diverges as $t →∞$, unless $y_0=y_{eq}$ b) As $t →∞$, the behavior of y depends upon the initial conditions (the value of $y_0$). c) There are three possible behaviors of $y$: If $y_0=-\frac{1}{2}$, then the value of y does not change; $y=-\frac{1}{2}$ for all $t$. As $t →∞$, $y→-\frac{1}{2}$. If $y_0\gt-\frac{1}{2}$, then as $t →∞$, $y→∞$. If $y_0\lt-\frac{1}{2}$, then as $t →∞$, $y→-∞$.

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 $y′=a+by$, 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) Pick several starting points: some below, some above, and some along $y_{eq}$. Trace an approximate graph of each particular $y$, based on the values of $y'$that were calculated, and observe the behavior of each 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. c)The behavior of the function $y(t)$ depends on $y_0$. If $y_0=-\frac{1}{2}$, then initial $y'=0$. The rate of change in $y$ is zero, so $y_0=y$ will be for all values of $t$. If $y_0\gt-\frac{1}{2}$, then as $t →∞$, $y→∞$. For all of these values of $y$, $y'\gt0$. Because of this, as $t$ increases, $y$ will continuously increase. If $y_0\lt-\frac{1}{2}$, then as $t →∞$, $y→-∞$. For all of these values of $y$, $y'\lt0$. Because of this, as $t$ increases, $y$ will continuously decrease.