Homework 1b

Prerequisite: The derivative of a function evaluated at a point = slope of the tangent line at that point.
This homework set contains two problems, each with multiple parts. Problem 1 is worth 4 points, while problem 2 is worth 1 point.

Direction fields consist of small portions of tangent lines drawn at several points.

For example, the direction field for $y' = (y - 1)(y +2)$ is drawn below.

Direction fields can be used to draw the graphs of solutions to differential equations. Several solutions are drawn on the direction field below.

Notice how these solutions stay tangent to the tangent lines. When drawing a solution starting at a point $(a, b)$, you can think of the path a boat would take if placed at this point and if the tangent lines indicated the direction of water flow.

A solution to a differential equation is an equilibrium solution if it is a constant solution, $y = C$. The graph of $y = C$ is a horizontal line. Observe in the direction field above that $y' = (y - 1)(y +2)$ has two equilibrium solutions $y = 1$ and $y = 2$.

If $y = C$ is a solution, then it should satisfy the differential equation when we plug it in. In particular, if $y = C$, then $y' = 0$. Thus to determine if $y' = f(t, y)$ has an equilibrium solution, plug in $y' = 0$ and solve for constant solutions (if any exist).

Most differential equations do not have equilibrium solutions; however, we will mostly look at differential equations of the form $y' = f(y)$ where the slope $y'$ depends only on $y$. Many, but not all, of these will have equilibrium solutions.

To summarize:

HW Problem 1: Draw direction fields by hand for the following differential equations. For these questions, $y'= f(y)$, and thus the slope only depends on $y$. Thus if the slope of the tangent line at $(t_0, y_0)$ is $m$, then the slope at $(t, y_0)$ is also $m$ for all values of $t$. Thus if you draw a small portion of a tangent line at $(t_0, y_0)$, you can quickly draw several other ones with the same slope at $(t, y_0)$ for various values of $t$ (but at the same height $y_0$). For each of the following problems:

HW Problem 1a. $~~~y' = y$
HW Problem 1b. $~~~y' = y(y-2)$
HW Problem 1c. $~~~y' = y(2-y)$
HW Problem 1d. $~~~y' = y(2-y)^2$

HW Problem 2: In calculus, you solved differential equations of the form $y' = f(t)$. We will look at one such example, so that you can compare the above problems to one from calculus.

HW Problem 2a: Draw the direction field for $y' = t$. Note this direction field does not have an equilibrium solution. If you plug in $y' = 0$, you get $t=0$, but this is not a horizontal line, $y = C$, and is thus not as equilibrium solution. If $t = 0$, the slope is zero, but these slopes occur along the $y-$axis (where $t = 0$). Thus you cannot draw a horizontal line connecting these slopes. Draw several slopes along the $y-$axis as well as several slopes for positive and negative $t$ values. Note the slopes depend only on $t$ and not $y$ for this calculus problem.

HW Problem 2b: Draw several solutions to the differential equation $y' = t$ on the direction field you drew for problem 2a. Note if $\frac{dy}{dt} = t$, then $dy = t dt$. Since this is a calculus problem, we can integrate both sides to determine $y$: $~~~~~y = \frac{1}{2}t^2 + C$. Observe that the solutions that you drew are of the form : $y = \frac{1}{2}t^2 + C$ where each solution is shifted vertically from the other solutions. This is very different than the solutions you drew in problem 1. Thus solutions for differential equations of the form $y' = f(t, y)$ can look very different than calculus 1 differential equations $y' = f(t)$. However, when you solve any first order differential equation, one constant $C$ will appear in the solution, but not usually as a simple shift of $+C$ as you had in calculus 1.

Direction fields can look very interesting. Below are some additional examples.

The differential equation $y' = y^2 + 1$ does not have an equilibrium solution:

The differential equation $y' = sin(t)sin(y)$ has infinitely many equilibrium solutions:

The direction field for $y' = t/y$ is drawn below. Observe that the solutions to $y' = t/y$ are semi-circles. In this class, we will assume that our solutions are always functions and thus satisfy the vertical line test. Thus a solution to $y' = t/y$ corresponds to either the top half or the botton half of a circle.

*Optional*: You can draw your own direction fields using the code in this notebook

To summarize:

Warning: **Direction fields can be misleading.** We will discuss this in sections 2.4 and 2.7.