Explanation Given: The given differential equation is x d y d x − 2 x 2 y + 3 x 2 = 0 . The provided initial condition is y ( 0 ) = 15 . Concepts used: Integrating factor: The function I ( x ) = e ∫ p ( x ) d x is called an integrating factor for the differential equation d y d x + P ( x ) y = Q ( x ) . Calculation: Consider the differential equation as x d y d x − 2 x 2 y + 3 x 2 = 0 (1) Compare above differential with d y d x + P ( x ) y = Q ( x ) . So, get the values P ( x ) = − 2 x and Q ( x ) = − 3 x . Now, find the integrating factor as I ( x ) = e ∫ p ( x ) d x which gives, I ( x ) = e − 2 ∫ x d x = e − x 2 Multiply each term of equation (1) by I ( x ) e − x 2 d y d x − 2 x e − x 2 y = − 3 x e − x 2 The sum of terms on the left can now be replaced with D x ( y ⋅ e − x 2 ) using the theory of differentials, D x ( y ⋅ e − x 2 ) = − 3 x e − x 2 Integrating both the sides of above equation gives y ⋅ e − x 2 = ∫ − 3 x e − x 2 d x Simplify, the above integral, y ⋅ e − x 2 = ∫ − 3 x e − x 2 d x (2) Now, to evaluate the integral on the right hand side Assume x 2 = t Differentiate both sides of above equation, 2 x d x = d t x d x = d t 2 Now put this value to evaluate the same integral,

Calculus with Applications Books a la Carte Edition

11th Edition

ISBN: 9780133864564

Author: Margaret L. Lial; Nathan P. Ritchey; Raymond N. Greenwell

Publisher: Pearson Education

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Differential Equations

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Chapter 10, Problem 45RE

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To solve: The given differential equation subject to given initial condition.

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Chapter 10, Problem 44RE

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Chapter 10 Solutions

Calculus with Applications Books a la Carte Edition

Ch. 10.1 - Find all solutions of the differential equation . Ch. 10.1 - Prob. 2YT Ch. 10.1 - Prob. 3YT Ch. 10.1 - Prob. 4YT Ch. 10.1 - Prob. 1WE Ch. 10.1 - Prob. 2WE Ch. 10.1 - Prob. 3WE Ch. 10.1 - Prob. 4WE Ch. 10.1 - Prob. 5WE Ch. 10.1 - Find the general solution for each differential...

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