We have found the following particular solution and its derivatives. = Ax² + Bx + C + (DX + E)ex Yp = 2AX + B + (Dx + E)ex + Dex = 2A + (DX + E)ex + 2Dex "p Substituting into the original differential equation results in the following. y" - 8y' + 20y = 200x278xex (2A + (DX + E)ex + 2Dex) − 8(2Ax + B + (Dx + E)ex + Dex) + 20(Ax² + Bx + C + (Dx + E)ex) = 200x² - 78xex Simplifying the left side of this equation gives the following. (2A + (Dx + E)ex + 2Dex) - 8(2Ax + B + (Dx + E)e* + Dex) + 20(Ax² + Bx + C + (Dx + E)ex) = (2A - 8B + 20C) + (-16A + 20B)x+ (-6D + 13E)ex + AX² As the coefficients of the terms in this simplified expression must be equal to the coefficients of 200x278xex, we have the following system. 2A8B20C = 0 -16A + 20B = 0 -6D + 13E = 0 = -78 = 200 Dxex +
We have found the following particular solution and its derivatives. = Ax² + Bx + C + (DX + E)ex Yp = 2AX + B + (Dx + E)ex + Dex = 2A + (DX + E)ex + 2Dex "p Substituting into the original differential equation results in the following. y" - 8y' + 20y = 200x278xex (2A + (DX + E)ex + 2Dex) − 8(2Ax + B + (Dx + E)ex + Dex) + 20(Ax² + Bx + C + (Dx + E)ex) = 200x² - 78xex Simplifying the left side of this equation gives the following. (2A + (Dx + E)ex + 2Dex) - 8(2Ax + B + (Dx + E)e* + Dex) + 20(Ax² + Bx + C + (Dx + E)ex) = (2A - 8B + 20C) + (-16A + 20B)x+ (-6D + 13E)ex + AX² As the coefficients of the terms in this simplified expression must be equal to the coefficients of 200x278xex, we have the following system. 2A8B20C = 0 -16A + 20B = 0 -6D + 13E = 0 = -78 = 200 Dxex +
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![We have found the following particular solution and its derivatives.
Ax² + Bx + C + (Dx + E)ex
= 2AX + B + (Dx + E)ex + Dex
= 2A + (Dx + E)ex + 2Dex
Ур
Yo
P
=
Yp
Substituting into the original differential equation results in the following.
y" - 8y' + 20y = 200x² - 78xex
(2A + (Dx + E)ex + 2Dex) − 8(2Ax + B + (Dx + E)ex + Dex)
+
Simplifying the left side of this equation gives the following.
(2A + (Dx + E)ex + 2De*) − 8(2Ax + B + (Dx + E)ex + Dex)
+ 20(Ax² + Bx + C + (Dx + E)ex)
= (2A - 8B + 20C) + (−16A + 20B)x + (−6D + 13E)ex +
])pxex + (1
AX²
As the coefficients of the terms in this simplified expression must be equal to the coefficients of 200x² - 78xex, we have the following system.
2A8B20C = 0
-16A + 20B = 0
-6D + 13E = 0
= -78
20(Ax² + Bx + C + (Dx + E)ex) = 200x² - 78xe*
A = 200](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc962edc4-7a48-4156-81ee-04d2df56a76e%2F11a8fe73-fa8b-4336-9d1a-19fa02713fad%2Fw7l92ic_processed.png&w=3840&q=75)
Transcribed Image Text:We have found the following particular solution and its derivatives.
Ax² + Bx + C + (Dx + E)ex
= 2AX + B + (Dx + E)ex + Dex
= 2A + (Dx + E)ex + 2Dex
Ур
Yo
P
=
Yp
Substituting into the original differential equation results in the following.
y" - 8y' + 20y = 200x² - 78xex
(2A + (Dx + E)ex + 2Dex) − 8(2Ax + B + (Dx + E)ex + Dex)
+
Simplifying the left side of this equation gives the following.
(2A + (Dx + E)ex + 2De*) − 8(2Ax + B + (Dx + E)ex + Dex)
+ 20(Ax² + Bx + C + (Dx + E)ex)
= (2A - 8B + 20C) + (−16A + 20B)x + (−6D + 13E)ex +
])pxex + (1
AX²
As the coefficients of the terms in this simplified expression must be equal to the coefficients of 200x² - 78xex, we have the following system.
2A8B20C = 0
-16A + 20B = 0
-6D + 13E = 0
= -78
20(Ax² + Bx + C + (Dx + E)ex) = 200x² - 78xe*
A = 200
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