Exploring with Slope Fields

1. Solutions, non-solutions, and Points

Below is a slope field for the first order differential equation $y' = -y + t$ . Not all equations are solvable explicitly, but this one is. Its solution family is

y = Ce^{-t}+t-1.

You will see two examples and non-examples for solutions curves.

For the question below, find the solution that matches the initial condition by dragging the curve's point to the initial condition point.

2 Integration Constant

It's not always as easy as dragging one point to another. Consider a familiar differential equation (because it's integration). The family of solutions for $y' = \cos(\frac{t}{2})$ is

y = 6 \sin\left(\frac{t}{2}\right) + C,

where $C$ is the $y$ -intercept that moves our function up and down.

Move the y-intercept up or down until you match the initial condition to find the right value for C.

What is the value of C that matches the initial condition y(9.5)=-1?

3 The Right Parameter

Below is a slope field for the first order differential equation $y' = y + t$ . Not all equations are solvable explicitly, but this one is. Its solution family is

y = Ce^t-t-1.

This time, $C$ is not the $y$ -intercept of the function, so instead a slider is given to control what $C$ is.

Use the slider below to find the right value of C for the solution that satisfies the initial condition.

What is the value of C that matches the initial condition y(4)=5?

4 Parameter Scale

Below is a slope field for the first order differential equation $y' = -2ty + 2t$ . Its solution family is

y = 1 + Ce^{-t^2}.

In this case, a very large value of $C$ is required to satisfy the initial condition below. Rather than asking you to move a slider to a higher value than Avogadro's number, you'll use a parameter D that scales C with a double exponential relationship:

C = e^{e^D}.

Use the slider below to find the right value of D for the solution that satisfies the initial condition.

What is the value of D that matches the initial condition y(9)=2?