Explanation Formula used: The integral of sine function is: ∫ e x = e x + C Calculation: The given ODE is y ′ + x e − x 2 2 = 0 . Rewrite the function as, y ′ = − x e − x 2 2 . This implies, d y d x = − x e − x 2 2 d y = − x e − x 2 2 d x Let u = − x 2 2 . This implies d u = − 2 x d x 2 = − x d x . Integrate and find the solution of ODE as follows

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ISBN: 9781119664697

Author: Kreyszig

Publisher: WILEY

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Chapter 1.1, Problem 2P

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The solution of ODE by integration or by differentiation y′+xe−x22=0.

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Chapter 1.1, Problem 1P

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Chapter 1 Solutions

Name: The solution of ODE by integration or by differentiation y ′ + x e − x 2 2 = 0 .
Uploaded: 2022-01-27
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Description: Explanation Formula used: The integral of sine function is: ∫ e x = e x + C Calculation: The given ODE is y ′ + x e − x 2 2 = 0 . Rewrite the function as, y ′ = − x e − x 2 2 . This implies, d y d x = − x e − x 2 2 d y = − x e − x 2 2 d x Let u = − x 2 2 . This implies d u = − 2 x d x 2 = − x d x . Integrate and find the solution of ODE as follows

ADVANCED ENGINEERING MATHEMATICS

Ch. 1.1 - Prob. 1P Ch. 1.1 - Prob. 2P Ch. 1.1 - Prob. 3P Ch. 1.1 - Prob. 4P Ch. 1.1 - Prob. 5P Ch. 1.1 - Prob. 6P Ch. 1.1 - Prob. 7P Ch. 1.1 - Prob. 8P Ch. 1.1 - Prob. 9P Ch. 1.1 - Prob. 10P

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