Explanation Procedure used: Particular solution to a linear DE by Green’s function: 1. Consider the differential equation and determine the complementary function to the homogeneous differential equation. 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. The general solution to a nonhomogeneous differential equation is given by y ( x ) = y h + y p , where, y h is the solution to the homogeneous equation associate to the differential equation L ( x ) with nonhomogeneous boundary conditions and y p is the solution to the nonhomogeneous differential equation with homogeneous boundary values. Calculation: Consider the differential equation y ″ + 2 y ′ + y = e − x . The associate homogeneous differential equation to y ″ + 2 y ′ + y = e − x is y ″ + 2 y ′ + y = 0 . The auxiliary equation to y ″ + 2 y ′ + y = 0 is m 2 + 2 m + 1 = 0 . Thus, calculate the solution to the auxiliary equation as, y ″ + 2 y ′ + y = 0 m 2 + 2 m + 1 = 0 ( m + 1 ) 2 = 0 Hence, the roots of the auxiliary equation are m 1 = − 1 and m 2 = − 1 . Thus, the two solutions to the DE are y 1 = e − x and y 2 = x e − x . Thus, the solution to the homogeneous equation is y h = c 1 e − x + c 2 x e − x . Since, there are no boundary values or initial values, leave the values of coefficients unknown. 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. Thus, y 1 ( x ) = e − x defines y 1 ( t ) = e − t , y 2 ( x ) = x e − x defines y 2 ( t ) = t e − t and and f ( x ) = e − x defines f ( t ) = e − t

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

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Chapter 3.10, Problem 9E

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The particular solution of the differential equation y″+2y′+y=e−x in the integral form.

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Chapter 3.10, Problem 8E

Chapter 3.10, Problem 10E

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For the Big-M tableau (of a maximization LP and row0 at bottom and M=1000), Z Ꮖ 1 x2 x3 81 82 83 e4 a4 RHS 0 7 0 0 1 0 4 3 -3 20 0 -4.5 0 0 0 1 -8 -2.5 2.5 6 0 7 0 1 0 0 8 3 -3 4 0 -1 50 1 0 0 0-2 -1 1 4 0000 0 30 970 200 If the original value of c₁ is increased by 60, what is the updated value of c₁ (meaning keeping the same set for BV. -10? Having made that change, what is the new optimal value for ž?

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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The particular solution of the differential equation y ″ + 2 y ′ + y = e − x in the integral form.

Solution Summary: The author explains the particular solution of the differential equation y′′+2e-x in the integral form and the general solution to a nonhomogeneous equation.

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