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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. (A computer algebra system is recommended. Round your answers to five decimal places.) y' = 7 - 5Vy. Y(0) = 1 (a) with h= 0.1 y(0.5) = 1.63906 y(1.0) = 1.84387 y(1.5) 1.91706

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. (A computer algebra system is recommended. Round your answers to five decimal places.) y' = 7 - 5Vy. Y(0) = 1 (a) with h= 0.1 y(0.5) = 1.63906 y(1.0) = 1.84387 y(1.5) 1.91706

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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I need help solving y(1.5), y(2.5), and y(3.0) of h=.01

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. (A computer algebra system is recommended. Round your answers to five decimal places.) y' = 7 - 5Vy, y(0) = 1 (a) with h = 0.1 y(0.5) = 1.63906 y(1.0) = 1.84387 y(1.5) = 1.91706 y(2.0) = 1.94401 y(2.5) = 1.95403 y(3.0) = 1.95777 (b) with h = 0.05 y(0.5) = 1.61790 y(1.0) = 1.82980 y(1.5) 1.90947 y(2.0) = 1.94025 y(2.5) = 1.95226 y(3.0) = 1.95696 (c) with h = 0.025 y(0.5) = 1.60807 y(1.0) = 1.82295 y(1.5) = 1.90563 y(2.0) = 1.93829 y(2.5) = 1.95130 y(3.0) = 1.95651 (d) with h = 0.01 y(0.5) = 1.60239 y(1.0) = 1.81891 y(1.5) y(2.0) = 1.93708 y(2.5) = 1.95607 y(3.0) = 1.95612
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. (A computer algebra system is recommended. Round your answers to five decimal places.) y' = 7 - 5Vy, y(0) = 1 (a) with h = 0.1 y(0.5) = 1.63906 y(1.0) = 1.84387 y(1.5) = 1.91706 y(2.0) = 1.94401 y(2.5) = 1.95403 y(3.0) = 1.95777 (b) with h = 0.05 y(0.5) = 1.61790 y(1.0) = 1.82980 y(1.5) 1.90947 y(2.0) = 1.94025 y(2.5) = 1.95226 y(3.0) = 1.95696 (c) with h = 0.025 y(0.5) = 1.60807 y(1.0) = 1.82295 y(1.5) = 1.90563 y(2.0) = 1.93829 y(2.5) = 1.95130 y(3.0) = 1.95651 (d) with h = 0.01 y(0.5) = 1.60239 y(1.0) = 1.81891 y(1.5) y(2.0) = 1.93708 y(2.5) = 1.95607 y(3.0) = 1.95612
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