Direction Fields Examples 1

# Direction Fields Examples 1

Recall from the Direction Fields page that we can sometimes learn a lot about the solutions to a differential equation by drawing a direction field which is a graph that contains line segment slopes for which particular points of particular solutions pass through with those tangential slopes. We will now look at some examples of direction fields. There isn't really much of an explainable process in constructing a direction field apart from plugging in values of $y$ and plotting down the slopes though, so be sure to attempt each question before looking at the solution.

## Example 1

Draw direction fields for the differential equations $y' = 2y + 4$, $y = 2y - 4$, $y = -2y + 4$, and $y = -2y - 4$ and describe the behaviour of the solutions to these differential equations as $t \to \infty$.

The directional fields for the four differential equations above are:

$y' = 2y + 4$ $y' = 2y - 4$
$y' = -2y + 4$ $y' = -2y - 4$
• For $y' = 2y + 4$, as $t \to \infty$, the solutions of this differential equation diverge away from the equilibrium solution $y = -2$.
• For $y' = 2y - 4$, as $t \to \infty$, the solutions of this differential equation diverge away from the equilibrium solution $y = 2$.
• For $y' = -2y + 4$, as $t \to \infty$, the solutions of this differential equation converge to the equilibrium solution $y = 2$.
• For $y' = -2y - 4$, as $t \to \infty$, the solutions of this differential equation converge to the equilibrium solution $y = -2$.

## Example 2

Determine for which values $a, b \in \mathbb{R}$, $a \neq 0$ that the solutions to the differential equation $y' = ay + b$ converge to the equilibrium solution and diverge from the equilibrium solution $y = -\frac{b}{a}$.

From example 1 above, we notice that the solutions to $y' = 2y + 4$ and $y' = 2y - 4$ diverge from their respective equilibrium solutions. Notice that in both of these differential equations, $a > 0$. Furthermore, we notice that the solutions to $y' = -2y + 4$ and $y' = -2y - 4$ both converge to their respective equilibrium solutions. In both of these differential equations, $a < 0$.

We conjecture that if $a > 0$ then all solutions to $y' = ay + b$ apart from the equilibrium solution $y = -\frac{b}{a}$ diverge from $y = -\frac{b}{a}$ and if $a < 0$ then all solutions to $y' = ay + b$ converge to the equilibrium solution $y = -\frac{b}{a}$.

To see that, suppose that $a > 0$. Then $\lim_{y \to \infty} y' = \lim_{y \to \infty} ay + b = \infty$ (since $a > 0$), and $\lim_{y \to -\infty} y' = ay + b = -\infty$ (once again since $a > 0$). Therefore as $y$ increases away from the equilibrium solution, the slopes of the tangents of any particular solution approach $\infty$ and thus diverge from the equilibrium solution. Similarly, as $y$ decreases away from the equilibrium solution, the slopes of the tangents of any particular solution approach $-\infty$ which points away from the equilibrium solution.

The same argument can be used to show that if $a < 0$ then the solutions to $y' = ay + b$ converge to the equilibrium solution.

## Example 3

Draw direction fields for the differential equations $y' = y(y + 2)$, $y = y(y - 2)$, $y = -y(y + 2)$, and $y = -y(y - 2)$ and describe the behaviour of the solutions to these differential equations as $t \to \infty$.

$y' = y(y + 2)$ $y' = y(y - 2)$
$y' = -y(y + 2)$ $y' = -y(y - 2)$
• For $y' = y(y + 2)$, as $t \to \infty$, the solutions of this differential equation converge to $y = -2$ if $y < 0$ and diverge otherwise.
• For $y' = y(y - 2)$, as $t \to \infty$, the solutions of this differential equation converge to $y = 0$ if $y < 2$ and diverge otherwise.
• For $y' = -y(y + 2)$, as $t \to \infty$, the solutions of this differential equation converge to $y = 0$ if $y > -2$ and diverge otherwise.
• For $y' = -y(y - 2)$, as $t \to \infty$, the solutions of this differential equation converge to $y = 2$ if $y > 0$ and diverge otherwise.