Section 66 Suppose we are given a differential equation and initial condition Then we can approximate the solution to the differential equation by its linearization which is close enough in a short interval ID: 212564
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Slide1
Euler’s Method
Section 6.6Slide2
Suppose we are given a differential equation and initial condition:
Then we can approximate the solution to the differential equation
by its linearization (which is “close enough” in a short interval
about
x
).
0
(solution curve)
The basis of Euler’s
method is to “string
together” many
linearizations to
approximate a curve.Slide3
Now, let’s specify a new value
for the independent variable:
If
dx
is small, then we have a new linearization:
From the point , which lies
exactly
on the solution curve,
we have obtained the point , which lies very close to the
point on the solution curve.Slide4
Second Step:
We use the point and the slope
of the solution curve through .
Setting , we use the linearization of the solution
curve through to calculate
This gives us the next approximation to values along the
solution curve .
Continue the pattern to find the third approximation:
Let’s see this process
graphically
…Slide5
Three steps in the Euler approximation to the solution of the
initial value problem
,
True solution curve
Euler approximation
ErrorSlide6
Using Euler’s Method
Find the first three approximations using Euler’s
method for the initial value problem
starting at with
We have:Slide7
Using Euler’s Method
Find the first three approximations using Euler’s
method for the initial value problem
starting at with
First ApproximationSlide8
Using Euler’s Method
Find the first three approximations using Euler’s
method for the initial value problem
starting at with
Second
ApproximationSlide9
Using Euler’s Method
Find the first three approximations using Euler’s
method for the initial value problem
starting at with
Third
ApproximationSlide10
Using Euler’s Method
Use Euler’s method to solve the given initial value problem on
the interval starting at with .
Compare the approximations to the values of the exact
solution.
Let me show you a new calculator program!!!Slide11
x
y
(Euler)
y
(exact)
Error
0
–2
–2
0
0.1
–1.8000
–1.8048
0.0048
0.2
–1.6100
–1.6187
0.0087
0.3
–1.4290
–1.4408
0.0118
0.4
–1.2561
–1.2703
0.0142
0.5
–1.0905
–1.1065
0.0160
0.6
–0.9314
–0.9488
0.0174
0.7
–0.7783
–0.7966
0.0183
0.8
–0.6305
–0.6493
0.0189
0.9
–0.4874
–0.5066
0.0191
1.0
–0.3487
–0.3679
0.0192Slide12
Other Practice Problems
Use differentiation and substitution to show that the given
function is the exact solution of the given initial value problem.
Initial Condition:
The same!
The same!Slide13
Other Practice Problems
Use analytic methods to find the exact solution of the given initial
v
alue problem.Slide14
Other Practice Problems
Use analytic methods to find the exact solution of the given initial
v
alue problem.
Initial Condition:
Solution:
orSlide15
Other Practice Problems
Use analytic methods to find the exact solution of the given initial
v
alue problem.
Initial Condition:
Solution: