#### CHECK OUT SHMOOP'S FREE STUDY TOOLS: Essay Lab | Math Shack | Videos

# Differential Equations

# Slope Fields and Solutions

If we have a slope field for a d.e. and a point in that slope field, we can sketch the solution that goes through that point. The little bits of tangent lines are like arrows telling the function which way to go. If we follow the arrows, we get the solution.

### Sample Problem

Here's the slope field for the d.e.

and a point:

Draw the solution that passes through the point.

Answer. The little tangent lines trace out the shape of the solution:

When we use slope fields to sketch solutions to differential equations, the solutions won't necessarily be functions.

### Sample Problem

Here's the slope field for the d.e.

and a point:

This isn't a function because it fails the vertical line test.

This is cool, because even if we can't get a single-variable function that's a solution to a differential equation, we can still see the shape of the solution on a graph.

Slope fields are the visual equivalent of IVPs for first-order differential equations. A slope field is the visual equivalent of a differential equation, and a point is the visual equivalent of an initial condition.

We can draw the solution that goes through a point, given a slope field and a point because there's a **Uniqueness Theorem** that says such a problem has exactly one answer.

Here's the fancy math language, written as bad poetry:

Given a first-order differential equation and an initial condition

(so long as the derivative is continuous)

there is exactly one solution that satisfies both the differential equation and the initial condition.

Here's the translation for slope fields:

Given a slope field and a point

(so long as the derivative is continuous)

there is exactly one solution that

goes through that point and has the shape specified by the slope field.

Since it's very unlikely that you'll ever be given such a problem where the derivative isn't continuous, you can probably get away with thinking that an IVP involving a first-order differential equation has exactly one solution.