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Transcript
```Technology for Chapter 11 and 12
Modeling with Differential Equations
Modeling with Systems of Differential Equations
Although much of the technology that we illustrate can solve more than just numerical
solutions, we will restrict our technology to numerical solutions, specifically Euler’s
Methods.
Chapter 11: Modeling with ODEs
Euler’s Algorithm:
Given the differential equation:
dy
 g (t , y ), y (t 0 )  y 0 , t0 < t < b
dt
Euler’s Method
Step1. Pick step size, h so that the interval, [b-t0]/n = h, is divided evenly.
Step 2. Start at y(t0)=y0.
Step 3. Compute yn+1 = yn + h* g( tn, yn)
Step 4. Let tn+1= tn + h, n=0, 1,2,…
Step 5. Continue until tn = b.
STOP
Lab example:
dy
 .05  (65  y ), y (0)  180
dt
0  t  30
We would like to be able to obtain numerical solution and their plots that look similar to these:
Maple:
dy
 .05  (65  y ), y (0)  180
dt
0  t  30
Use a step size of h=1
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How did we do?
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Euler’s method appears to have done a good job in the approximations.
Chapter 12: Modeling with Systems of ODES
Euler’s method:
Consider the iterative formula for Euler’s Method for systems as
x(n)=x(n-1)+f(t(n-1),x(n-1),y(n-1))t
y(n)=y(n-1)+g(t(n-1),x(n-1),y(n-1))t
We illustrate a few iterations for the following initial value problem with a step size of t=0.1 :
x’=3x-2y, x(0)=3
y’=5x-4y, y(0)=6
x(0)=3, y(0)==6 Given
x(1)  3  (0.1)  (3  3  2  6)  2.7
y (1)  6  (0.1)  (5  3  4  6)  5.1
and
x(2)  2.7  (0.1)  (3  (2.7)  2  (5.1))  2.49
y (2)  5.1  (0.1)  (5  (2.7)  4  (5.1))  4.41
and so forth.
We complete in Maple and illustrate below.
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