Transcript AP CALCULUS AB

AP CALCULUS AB
Chapter 6:
Differential Equations and
Mathematical Modeling
Section 6.1:
Slope Fields and Euler’s Method
What you’ll learn about
Differential Equations
 Slope Fields
 Euler’s Method

… and why
Differential equations have been a prime
motivation for the study of calculus and
remain so to this day.
Differential Equation
An equation involving a derivative is
called a differential equation. The
order of a differential equation is
the order of the highest derivative
involved in the equation.
Example Solving a Differential
Equation
Find all functions y that satisfy
dy
 3 x  cos x.
dx
2
The solution can be any antiderivative of 3x2  cos x, which can be any
function of the form y  x3  sin x  C.
First-order Differential Equation
If the general solution to a first-order differential
equation is continuous, the only additional
information needed to find a unique solution is
the value of the function at a single point, called
an initial condition. A differential equation
with an initial condition is called an initial-value
problem. It has a unique solution, called the
particular solution to the differential equation.
Section 6.1 – Slope Fields and Euler’s
Method

Example: Solve
the differential
equation
dy
2
 3x  3
dx
for an initial
condition
that y = 2 when
x = 1.
dy
 3x 2  3
dx
dy   3 x 2  3  dx
2
dy

3
x
    3 dx
y  x3  3x  C
2  13  3 1  C
Solution to the
Differential
Equation.
2  4C
2  C
y  x3  3x  2
Solution to the
Initial value
problem
Section 6.1 – Slope Fields and Euler’s
Method
Example 2:
Differential
equation:

dy
 2y
dx
Initial condition:
y  0  12
dy
 2y
dx
dy  2 ydx
1
dy  2dx
y
1
 y dy   2dx
ln y  2 x  C
C1e
2 0 
 12
C1 1  12
C1  12
e 2 x C  y
Solution :
e 2 x eC  y
y  12e2 x
C1e 2 x  y
Example Solving an Initial Value
Problem
dy
Find the particular solution to the equation
 e2 x  3x
dx
 1
whose graph passes through the point 1,  .
 2
1 2x 3 2
e  x  C.
2
2
Applying the initial condition, we have
The general solution is y 
1 1 2 3
 e   C , from which we conclude that
2 2
2
1
C  2  e 2 . Therefore, the particular equation is
2
1
3
1
y  e2 x  x 2  2  e2 .
2
2
2
Example Using the Fundamental Theorem to Solve
an Initial Value Problem
Find the solution to the differential equation
f '( x)  cos  x2  for which f (3)  5.
The solution is f (x)   cos  t  dt  5.
x
2
3
For x=3, the integral is 0+5 (i.e. this particular curve is
translated vertically 5 units, with no thickness if we start
the integral at 3.
Section 6.1 – Slope Fields and Euler’s
Method

A slope field or
direction field for the
first order differential
equation
dy
 f  x, y 
dx
is a plot of short line
segments with slopes
f(x, y) for a lattice of
points (x, y) in the
plane.
Example Constructing a Slope
Field
Construct a slope field for the differential equation
dy
 cos x.
dx
The slope at any point  0, y  will be cos 0  1.
The slope at any point  , y  or   , y  will be -1.
The slope at all odd multiples of

will be 0.
2
The slope is 1 along the lines x  2 .
Section 6.1 – Slope Fields and Euler’s
Method

Example: dy  2 x, y 0   2
dx
To plot the slope
field for this
differential equation,
plug in values for x
and plot short lines
to represent slope
 Then use your initial
value to determine
the actual curve.

Section 6.1 – Slope Fields and Euler’s
Method

Example (cont.)
dy
 2 x, y 0   2
dx
dy  2 xdx
 dy   2 xdx
y  x2  C
 2  02  C
2C
y  x2  2
Euler’s Method for Graphing a
Solution to an Initial Value Problem
1. Begin at the point ( x, y ) specified by t he initial condition.
2. Use the differential equation to find the slope dy / dx at the point.
3. Increase x by Dx . Increase y by D y , where
D y (dy / dxD) x. This defines a new point (x  Dx, y  Dy)
that lies along the linearization.
4. Using this new point, return to step 2. Repeating the
process constructs the graph to the right of the initial point.
5. To construct the graph
moving to the left from the
initial point, repeat the process
using negative values for D x.
Example Applying Euler’s
Method
Use Euler's Method with increments of Dx  0.1
to approximate the value of y when x 1.5
given dy  x 1 and y  2 when x 1.
dx
 x, y 
Dx
 dy 
Dy    Dx  x  Dx, y  Dy 
 dx 
(1, 2)
dy
 x 1
dx
2
0.1
0.2
(1.1, 2.2)
(1.1, 2.2)
2.1
0.1
0.21
(1.2, 2.41)
(1.2, 2.41)
2.2
0.1
0.22
(1.3, 2.63)
(1.3, 2.63)
2.3
0.1
0.23
(1.4, 2.86)
(1.4, 2.86)
2.4
0.1
0.24
(1.5, 3.1)
Euler's Method leads us to the approximation f (1.5)  3.1.