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 initial 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. 4. Obtain the final solution by y ( x ) = y h + y p . Calculation: The given initial-value problem is y ″ + 6 y ′ + 9 y = x , y ( 0 ) = 1 , y ′ ( 0 ) = − 3 . Consider the differential equation y ″ + 6 y ′ + 9 y = x . The associated homogeneous differential equation is y ″ + 6 y ′ + 9 y = 0 . The auxiliary equation to y ″ + 6 y ′ + 9 y = 0 is m 2 + 6 m + 9 = 0 . Calculate the solution to the auxiliary equation, m 2 + 6 m + 9 = 0 ( m + 3 ) 2 = 0 Hence, the roots of the auxiliary equation are m 1 = − 3 and m 2 = − 3 . Thus, the two solutions to the DE are y 1 = e − 3 x and y 2 = x e − 3 x . Hence, the solution of the homogeneous DE y ″ + 6 y ′ + 9 y = 0 is y h ( x ) = c 1 e − 3 x + c 2 x e − 3 x . Consider the boundary values y ( 0 ) = 1 , y ′ ( 0 ) = − 3 , Substitute y ( 0 ) = 1 into y h ( x ) = c 1 e − 3 x + c 2 x e − 3 x , 1 = c 1 + c 2 c 1 = 1 − c 2 The first differential of y h ( x ) = c 1 e − 3 x + c 2 x e − 3 x is, y ′ h = − 3 c 1 e − 3 x + c 2 e − 3 x − 3 c 2 x e − 3 x Substitute the boundary value y ′ ( 0 ) = − 3 , − 3 = − 3 c 1 + c 2 − 3 = − 3 c 1 + 1 − c 1 − 4 c 1 = − 4 c 1 = 1 Hence, c 2 = 0 . Thus, the values of the coefficients are c 1 = 1 and c 2 = 0 . Hence, the solution to homogeneous DE y ″ + 6 y ′ + 9 y = 0 is y h = e − 3 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, to determine the Wronskian, use variable transformation, [ y 1 ( x ) = e − 3 x defines y 1 ( t ) = e − 3 t , y 2 ( x ) = x e − 3 x defines y 2 ( t ) = t e − 3 t and f ( x ) = x defines f ( t ) = t ] W ( t ) = | e − 3 t t e − 3 t − 3 e − 3 t − 3 t e − 3 t + e − 3 t | = − 3 t e − 6 t + e − 6 t + 3 t e − 6 t = e − 6 t Thus, for any t , W ( t ) ≠ 0 , determine the greens function G ( x , t ) , G ( x , t ) = y 1 ( t ) y 2 ( x ) − y 1 ( x ) y 2 ( t ) W ( t ) = e − 3 t x e − 3 x − e − 3 x t e − 3 t e − 6 t = x e − 3 x e 3 t − t e − 3 x e 3 t = e 3 ( t − x ) ( x − t ) Substitute the green function into y p = ∫ x 0 x G ( x , t ) f ( t ) d t , y p = ∫ 0 x e 3 ( t − x ) ( x − t ) ⋅ t d t

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3.10, Problem 22E

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The particular solution of the initial-value problem.

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Chapter 3.10, Problem 21E

Chapter 3.10, Problem 23E

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Here is the optimal tableau for a standard Max problem. zx1 x2 x3 24 81 82 83 rhs 1 0 5 3 0 6 0 1 .3 7.5 0 - .1 .2 0 0 28 360 0 -8 522 0 2700 0 6 12 1 60 0 0 -1/15-3 1 1/15 -1/10 0 2 Using that the dual solution y = CBy B-1 and finding B = (B-¹)-¹ we find the original CBV and rhs b. The allowable increase for b₂ is If b₂ is increased by 3 then, using Dual Theorem, the new value for * is If c₂ is increased by 10, then the new value for optimal > is i.e. if no change to BV, then just a change to profit on selling product 2. The original coefficients c₁ = =☐ a and c4 = 5 If c4 is changed to 512, then (first adjusting other columns of row0 by adding Delta times row belonging to x4 or using B-matrix method to update row0) the new optimal value, after doing more simplex algorithm, for > is

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Chapter 3 Solutions

Advanced Engineering Mathematics

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