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er's Method to find approximate values of the solution of the given initial value problem at t = 0.5, 1, 1.5, 2, 2.5, and y' = 7 - 4√y, y(0) = 3 (a) with h = 0.1 y(0.5) ≈ 0.06372 y(1.0) 0.0568 y(1.5) y(2.0) y(2.5) y(3.0) Z (b) with h = 0.05 y(0.5) y(1.0) y(1.5) y(2.0) y(2.5)= y(3.0) (c) with h= 0.025 y(0.5) y(1.0) y(1.5) y(2.0) y(2.5) y(3.0) ≈ (d) with h = 0.01 y(0.5) y(1.0) y(1.5) y(2.0) y(2.5) y(3.0) X X

er's Method to find approximate values of the solution of the given initial value problem at t = 0.5, 1, 1.5, 2, 2.5, and y' = 7 - 4√y, y(0) = 3 (a) with h = 0.1 y(0.5) ≈ 0.06372 y(1.0) 0.0568 y(1.5) y(2.0) y(2.5) y(3.0) Z (b) with h = 0.05 y(0.5) y(1.0) y(1.5) y(2.0) y(2.5)= y(3.0) (c) with h= 0.025 y(0.5) y(1.0) y(1.5) y(2.0) y(2.5) y(3.0) ≈ (d) with h = 0.01 y(0.5) y(1.0) y(1.5) y(2.0) y(2.5) y(3.0) X X

Advanced Engineering Mathematics
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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Use Euler's Method to find approximate values of the solution of the given initial value problem at 

t = 0.5, 1, 1.5, 2, 2.5, and 3. Round to 5 decimal places.
Use Euler's Method to find approximate values of the solution of the given initial value problem at t = 0.5, 1, 1.5, 2, 2.5, and 3. y' = 7 - 4√√y, y(0) = 3 (a) with h = 0.1 y(0.5) 0.06372 y(1.0) ≈ 0.0568 y(1.5) y(2.0) y(2.5) y(3.0) (b) with h = 0.05 y(0.5) y(1.0) y(1.5) y(2.0) y(2.5) y(3.0) ~ ~ (c) with h = 0.025 y(0.5) y(1.0) ~ y(1.5)≈ y(2.0) y(2.5) ~ y(3.0) (d) with h = 0.01 y(0.5) y(1.0): y(1.5) y(2.0) y(2.5) y(3.0) ≈
Transcribed Image Text:Use Euler's Method to find approximate values of the solution of the given initial value problem at t = 0.5, 1, 1.5, 2, 2.5, and 3. y' = 7 - 4√√y, y(0) = 3 (a) with h = 0.1 y(0.5) 0.06372 y(1.0) ≈ 0.0568 y(1.5) y(2.0) y(2.5) y(3.0) (b) with h = 0.05 y(0.5) y(1.0) y(1.5) y(2.0) y(2.5) y(3.0) ~ ~ (c) with h = 0.025 y(0.5) y(1.0) ~ y(1.5)≈ y(2.0) y(2.5) ~ y(3.0) (d) with h = 0.01 y(0.5) y(1.0): y(1.5) y(2.0) y(2.5) y(3.0) ≈
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