Answer the following items on a clean sheet of paper. Upload photos of your answer sheets showing your solutions. Df = 2x – f with f(0): f(2) :-1 1. Consider the ODE With step size 1, approximate using Euler's method.
Answer the following items on a clean sheet of paper. Upload photos of your answer sheets showing your solutions. Df = 2x – f with f(0): f(2) :-1 1. Consider the ODE With step size 1, approximate using Euler's method.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Transcribed Image Text:Answer the following items on a clean sheet of paper. Upload photos of your answer sheets showing your solutions.
Df = 2x – f with
f (0) = –1
f(2).
1. Consider the ODE
With step size 1, approximate
using Euler's method.
![Example 2.1
Consider the ODE Df – 2f + 3x = 0; f (0) = 1. With n = 2, approximate f (2)
a. Using Euler's Method.
b. Using Backward Euler's Method.
Solution:
We can rewrite the ODE as Df = 2f – 3x. Let g(x,f) = 2f – 3x. Dividing [0,2] into two (2) intervals, with
length h = = 1, we get xo = 0, x1 = 0 +1 = 1, and x, = 1+1= 2. Let yo = f (0) = 1.
2-0
2
a. Using Euler's Method,
b. Using Backward Euler's Method,
Yı = yo + 2yo – 3x,
= 1+ 2
= 3
Y2 = Yı + 2y1 – 3x1
= 3 + 2(3) – 3
= 6
Y1 = yo + 2y1 - 3x1
= 1+ 2y1 – 3
-Yı = -2
Y1 = 2
Y2 = y1 + 2y2 - 3x2
= 2 + 2y2 – 3(2)
-Y2 = -4
Y2 = 4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F639a7ef1-f4e6-4fee-a616-51c4d4cf8a40%2F2fa548dc-ce08-47e4-b89a-5f8955a62fb7%2F1r06lu_processed.png&w=3840&q=75)
Transcribed Image Text:Example 2.1
Consider the ODE Df – 2f + 3x = 0; f (0) = 1. With n = 2, approximate f (2)
a. Using Euler's Method.
b. Using Backward Euler's Method.
Solution:
We can rewrite the ODE as Df = 2f – 3x. Let g(x,f) = 2f – 3x. Dividing [0,2] into two (2) intervals, with
length h = = 1, we get xo = 0, x1 = 0 +1 = 1, and x, = 1+1= 2. Let yo = f (0) = 1.
2-0
2
a. Using Euler's Method,
b. Using Backward Euler's Method,
Yı = yo + 2yo – 3x,
= 1+ 2
= 3
Y2 = Yı + 2y1 – 3x1
= 3 + 2(3) – 3
= 6
Y1 = yo + 2y1 - 3x1
= 1+ 2y1 – 3
-Yı = -2
Y1 = 2
Y2 = y1 + 2y2 - 3x2
= 2 + 2y2 – 3(2)
-Y2 = -4
Y2 = 4
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