Explanation Procedure used: Particular solution to a linear DE by greens function. 1. Consider the differential equation and determine the complementary function to the homogeneous differential equation. Further, substitute the boundary values to find the coefficients. 2. If y 1 ( x ) and y 2 ( x ) are two fundamental solution of the differential equation then particular solution is given by y p = u 1 y 1 ( x ) + u 2 y 2 ( x ) . 3. Use, y p = ∫ x 0 x G ( x , t ) f ( t ) d t to find the particular solution. Find greens function from G ( x , t ) = y 1 ( t ) y 2 ( x ) W t . 4. Obtain the final solution by y ( x ) = y h + y p . Calculation: Consider the differential equation x 2 y ″ + x y ′ = 1 . The associated homogeneous differential equation is x 2 y ″ + x y ′ = 0 . Since the differential equation is a Cauchy-Euler equation. Thus, substitute y = x m into the homogeneous DE x 2 y ″ + x y ′ = 0 . x 2 m ( m − 1 ) y m − 2 + x m x m − 1 = 0 Since x m ≠ 0 for any m . Thus, m ( m − 1 ) + m = 0 m 2 − m + m = 0 m 2 = 0 Thus, the roots of the auxiliary equation are m 1 = 0 and m 2 = 0 Hence, the solution of homogeneous DE x 2 y ″ + x y ′ = 0 is y h = c 1 + c 2 ln x . Consider the boundary values y ( e − 1 ) = 0 , y ( 1 ) = 0 . Substitute y ( e − 1 ) = 0 into y h = c 1 + c 2 ln x . Thus, 0 = c 1 + c 2 ln e − 1 0 = c 1 − c 2 Thus, choose c 2 = 1 . Hence, y 1 = 1 + ln x . Substitute y ( 1 ) = 0 . 0 = c 1 + c 2 ln ( 1 ) Choose, c 2 = 1 . Thus, y 2 = ln x . The particular solution to the nonhomogeneous differential equation is given by y p = ∫ x 0 x G ( x , t ) f ( t ) d t . Now, determine the Wronskian. Use, variable transformation. Thus, [ y 1 ( x ) = 1 + ln x defines y 1 ( t ) = 1 + ln t , y 2 ( x ) = ln x defines y 2 ( t ) = ln t f ( x ) = 1 defines f ( t ) = 1 ] W ( t ) = | y 1 ( t ) y 2 ( t ) y ′ 1 ( t ) y ′ 2 ( t ) | = | 1 + ln t ln t 1 t 1 t | = 1 t + 1 t ln t − 1 t ln t = t − 1 Thus, for any t W ( t ) ≠ 0 . Determine the greens function G ( x , t ) . Thus, G ( x , t ) = y 1 ( t ) y 2 ( x ) W ( t ) [ For e − 1 ≤ t ≤ x here, a = e − 1 and b = 1 ] = ( 1 + ln t ) ln x t − 1 = ln x ( 1 + ln t ) t and G ( x , t ) = y 1 ( x ) y 2 ( t ) W ( t ) [ For x ≤ t ≤ 1 here, a = e − 1 and b = 1 ] = ( 1 + ln x ) ( ln t ) t − 1 = ( 1 + ln x ) ( t ⋅ ln t ) Divide the DE by x 2

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 4.8, Problem 43E

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The solution of the given boundary values problem $x2y″+xy′=1,y(e−1)=0,y(1)=0$ .

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Chapter 4.8, Problem 42E

Chapter 4.8, Problem 44E

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Differential Equations with Boundary-Value Problems (MindTap Course List)

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The solution of the given boundary values problem x 2 y ″ + x y ′ = 1 , y ( e − 1 ) = 0 , y ( 1 ) = 0 .

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The solution of the given boundary values problem $x2y″+xy′=1,y(e−1)=0,y(1)=0$ .

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