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Use Euler's method to approximate y(1.0). Start with step size h = 0.1, and then use successively smaller step sizes (h=0.01, 0.001, 0.0001, etc.) until successive approximate solution values at x = 1.0 agree rounded off to two decimal places. y' = x² + y² -1, y(0) = 0 ... The approximate solution values at x = 1.0 begin to agree rounded off to two decimal places between is. (Type an integer or decimal rounded to two decimal places as needed.) So, a good approximation of y(1.0) h = 0.001 and h = 0.0001. h = 0.1 and h=0.01. h = 0.01 and h = 0.001.

Use Euler's method to approximate y(1.0). Start with step size h = 0.1, and then use successively smaller step sizes (h=0.01, 0.001, 0.0001, etc.) until successive approximate solution values at x = 1.0 agree rounded off to two decimal places. y' = x² + y² -1, y(0) = 0 ... The approximate solution values at x = 1.0 begin to agree rounded off to two decimal places between is. (Type an integer or decimal rounded to two decimal places as needed.) So, a good approximation of y(1.0) h = 0.001 and h = 0.0001. h = 0.1 and h=0.01. h = 0.01 and h = 0.001.

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 approximate y(1.0). Start with step size h = 0.1, and then use successively smaller step sizes (h= 0.01, 0.001, 0.0001, etc.) until successive approximate solution values at x = 1.0 agree rounded off to two decimal places. y' = x² + y² -1, y(0) = 0 The approximate solution values at x = 1.0 begin to agree rounded off to two decimal places between is (Type an integer or decimal rounded to two decimal places as needed.) So, a good approximation of y(1.0) h = 0.001 and h = 0.0001. h = 0.1 and h=0.01. h = 0.01 and h = 0.001.
Transcribed Image Text:Use Euler's method to approximate y(1.0). Start with step size h = 0.1, and then use successively smaller step sizes (h= 0.01, 0.001, 0.0001, etc.) until successive approximate solution values at x = 1.0 agree rounded off to two decimal places. y' = x² + y² -1, y(0) = 0 The approximate solution values at x = 1.0 begin to agree rounded off to two decimal places between is (Type an integer or decimal rounded to two decimal places as needed.) So, a good approximation of y(1.0) h = 0.001 and h = 0.0001. h = 0.1 and h=0.01. h = 0.01 and h = 0.001.
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