(a) 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 y ″ = f ( x ) . The associate homogeneous differential equation to y ″ = f ( x ) is y ″ = 0 . The auxiliary equation to the homogeneous DE y ″ = 0 is m 2 = 0 . Find the roots of m 2 = 0 . Thus, the roots of the auxiliary equation are m 1 = 0 and m 2 = 0 Hence, the solution of homogeneous DE y ″ = 0 is y h ( x ) = c + c 2 x . Consider the boundary values y ( 0 ) = 0 , y ( 1 ) + y ′ ( 1 ) = 0 . Substitute y ( 0 ) = 0 into y h ( x ) = c + c 2 x . Thus, 0 = c 1 Thus, choose c 2 = 1 . Hence, y 1 = x . Substitute y ( 1 ) + y ′ ( 1 ) = 0 . Thus, 0 = c 1 + c 2 + c 2 2 c 1 = − c 1 Choose, c 2 = − 1 . Thus, y 2 = 2 − 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 ) = x defines y 1 ( t ) = t , y 2 ( x ) = 2 − x defines y 2 ( t ) = 2 − t for 0 ≤ < t ≤ 0 ] , and. W ( t ) = | y 1 ( t ) y 2 ( t ) y ′ 1 ( t ) y ′ 2 ( t ) | = | t 2 − t 1 − 1 | = − t − 2 + t = − 2 Thus, for any t W ( t ) ≠ 0 (b) To determine To verify: If the particular solution obtained verifies the boundary conditions y ( 0 ) = 0 , y ( 1 ) + y ′ ( 1 ) = 0 .

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ISBN: 9781337652469

Author: ZILL

Publisher: CENGAGE L

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Chapter 4.8, Problem 36E

(a)

To determine

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

(b)

To determine

To verify: If the particular solution obtained verifies the boundary conditions y(0)=0,y(1)+y′(1)=0.

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

Solution Summary: The author explains the solution of the boundary value problem y′′=f(x)

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

To verify: If the particular solution obtained verifies the boundary conditions y(0)=0,y(1)+y′(1)=0.

Chapter 4 Solutions