First, a little review:

Consider:2 3y x

then: 2y x 2y x

2 5y x or

It doesn’t matter whether the constant was 3 or -5, since when we take the derivative the constant disappears.

However, when we try to reverse the operation:

Given: 2y x find y

2y x C

We don’t know what the constant is, so we put “C” in the answer to remind us that there might have been a constant.

If we have some more information we can find C.

Given: and when , find the equation for .2y x y4y 1x

2y x C 24 1 C

3 C2 3y x

This is called an initial value problem. We need the initial values to find the constant.

An equation containing a derivative is called a differential equation. It becomes an initial value problem when you are given the initial condition and asked to find the original equation.

General and Particular Solutions

A function y = f(x) is called a solution of a differential equation if the equation is satisfied when y and its derivatives are replaced by f(x) and its derivatives.

y' + 2y = 0 y = 5e-2x

differential equation particular solutiongeneral solution

y = Ce-2x

y' + 2y = 0

The order of a differential equation is determined by the highest-order derivative in the equation.

First-order

Second-ordery'' = -32

Initial value problems and differential equations can be illustrated with a slope field.

Slope fields are mostly used as a learning tool and are mostly done on a computer or graphing calculator, but a recent AP test asked students to draw a simple one by hand.

Draw a segment with slope of 2.

Draw a segment with slope of 0.

Draw a segment with slope of 4.

2y x

x y y0 0 00 1 00 00 0

23

1 0 21 1 2

2 0 4

-1 0 -2

-2 0 -4

2y x

If you know an initial condition, such as (1,-2), you can sketch the curve.

By following the slope field, you get a rough picture of what the curve looks like.

In this case, it is a parabola.

Sketch the solution for the equation y' = 1/3 x2 - 1/2 xthat passes through (1, 1)

Sketch the solution for the equation y' = y + xythat passes through (0, 4)

y' = xy

Sketch the solution that passes through the point (0, 1/2)

y' = xy

Sketch the solution that passes through the point (0, 1/2)

The AP Board Expects you to be able to do the following with Slope Fields

• Sketch a slope field for a given differential equation

• Given a slope field, sketch a solution curve through a given point

• Match a slope field to a differential equation

• Match a slope field solution to a differential equation

Sketching a Slope Field•Compute the slope for each point

•Make a small mark that approximates the slope through the point.

1dy

xdx

When x = 0, dy/dx = -1

When x = 1, dy/dx = 0

When x = 2, dy/dx = 1

When x = -1, dy/dx = -2

When x = -2, dy/dx = -3

Sketching a Slope Field in x and ydy

x ydx

You may find it useful to make a table

-2 -1 0 1 2

3 -5 -4 -3 -2 -1

2 -4 -3 -2 -1 0

1 -3 -2 -1 0 1

0 -2 -1 0 1 2

-1 -1 0 1 2 3

-2 0 1 2 3 4

y

x

Example based on prior year free response questions

dy x

dx y

Graph the slope field.

-1 0 1

3 -1/3 0 1/3

2 -1/2 0 1/2

1 -1 0 1

0 ---- ---- ----

-1 1 0 -1

y

x

Let y = f(x) be the particular solution with the initial condition f(1) = 2

Write an equation of the tangent line to the curve at (1, 2).

We have the point (1, 2)

At (1, 2) we had a slope of 1/2

12 1

21 3

2 2

y x

y x

Use the line to approximate f(1.1) 1 3

1.1 2.052 2

y

Find the particular solution to y = f(x) with the initial condition f(1) = 2.

2 2

2 2

dy x

dx y

ydy xdx

ydy xdx

y xC

2 2

2 2

2

2 1

2 21

221.5

1.52 2

3

C

C

c

y x

y x

Plug in the point

Keep only the positive root. Why?