Explanation Given: The differential equation is d y d x + 2 x y = f ( x ) with the initial condition y ( 0 ) = 2 . where, f ( x ) = { x , 0 ≤ x < 1 0 , x ≥ 1 (1) Procedure used: The standard form of the first order linear differential equation with initial condition y ( x 0 ) = y 0 is, d y d x + P ( x ) y = Q ( x ) (2) To solve the equation (2) for y ( x ) , use the integrating factor method. The integrating factor of the differential equation is e ∫ P ( x ) d x . And the solution of the differential equation is, y ⋅ ( Integrating factor ) = ∫ ( ( Integrating factor ) ⋅ ( Q ( x ) ) ) d x + C (3) Here, C is constant. Calculation: From equation (1), the differential equation on the interval ( 0 , 1 ) with initial condition y ( 0 ) = 2 is, d y d x + 2 x y = x (4) Compare equation (4) with equation (2). P ( x ) = 2 x and Q ( x ) = x Therefore the integrating factor of the differential equation (4) is, I . F = e ∫ 2 x d x = e x 2 Substitute e x for integrating factor, 1 for Q ( x ) , and C 1 for C in equation (3). y ⋅ e x 2 = ∫ ( e x 2 ⋅ x ) d x + C 1 y ⋅ e x 2 = ∫ ( x e x 2 ) d x + C 1 (5) Use the method of substitution, and further simplify. x 2 = u 2 x d x = d u x d x = d u 2 Substitute d u 2 for x d x and u for x 2 in equation (5). y ⋅ e u = ∫ ( e u ⋅ d u 2 ) + C 1 y ⋅ e u = e u 2 + C 1 (6) Substitute x 2 for u in equation (6). y ⋅ e x 2 = e x 2 2 + C 1 (7) Use the initial condition y ( 0 ) = 2 in equation (7). 2 × e 0 = e 0 2 + C 1 2 = 1 2 + C 1 C 1 = 2 − 1 2 C 1 = 3 2 Substitute 3 2 for C 1 in equation (7). y ⋅ e x 2 = e x 2 2 + 3 2 y = 1 e x 2 ( e x 2 2 + 3 2 ) y = 1 2 + 3 2 e x 2 y = 1 2 + 3 e − x 2 2 (8) From equation (1), the differential equation on the interval ( 1 , ∞ ) with initial condition y ( 0 ) = 2 is, d y d x + 2 x y = 0 (9) Compare equation (9) with equation (1)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 2.3, Problem 35E

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Chapter 2.3, Problem 34E

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