5.5.4 Example D Let us investigate the case where the inhomogeneous term contains a factor that also appears in the solution of the homogeneous equation. Consider the first-order equation (aE1 + bE2 + c)z(k, l) = F(k, (), (5.183) where a, b, and c are constants, and F(k, l) satisfies the condition (aE1 + bE2 + c)F(k,l) = 0. (5.184) 188 Difference Equations If we assume that the particular solution has the form Zp(k, l) = (Ak + Bl)F(k, l), (5.185) where A and B are unknown constants, then substitution of this result into equation (5.183) gives (AAE1 + 6BE2 – 1)F(k, l) = 0. (5.186) Comparison of equations (5.184) and (5.186) allows the conclusion A = B = -1/c. (5.187) Therefore, for this case, the particular solution is Zp(k, l) = (-1/c)(k + l)F(k,l). (5.188)
5.5.4 Example D Let us investigate the case where the inhomogeneous term contains a factor that also appears in the solution of the homogeneous equation. Consider the first-order equation (aE1 + bE2 + c)z(k, l) = F(k, (), (5.183) where a, b, and c are constants, and F(k, l) satisfies the condition (aE1 + bE2 + c)F(k,l) = 0. (5.184) 188 Difference Equations If we assume that the particular solution has the form Zp(k, l) = (Ak + Bl)F(k, l), (5.185) where A and B are unknown constants, then substitution of this result into equation (5.183) gives (AAE1 + 6BE2 – 1)F(k, l) = 0. (5.186) Comparison of equations (5.184) and (5.186) allows the conclusion A = B = -1/c. (5.187) Therefore, for this case, the particular solution is Zp(k, l) = (-1/c)(k + l)F(k,l). (5.188)
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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Explain the determine blue
Transcribed Image Text:5.5.4 Example D
Let us investigate the case where the inhomogeneous term contains a factor
that also appears in the solution of the homogeneous equation. Consider the
first-order equation
(aE1 + bE2 + c)z(k, l) = F(k,l),
(5.183)
where
a,
b, and c are constants, and F(k, l) satisfies the condition
(aE1 + bE2 + c)F(k, l) = 0.
(5.184)
188
Difference Equations
If we assume that the particular solution has the form
Zp(ki, l) = (Ak + Bl)F(k, l),
(5.185)
where A and B are unknown constants, then substitution of this result into
equation (5.183) gives
(AAE1 + 6BE2 –- 1)F(k, l) = 0.
(5.186)
Comparison of equations (5.184) and (5.186) allows the conclusion
A = B = -1/c.
(5.187)
Therefore, for this case, the particular solution is
Zp(k, l) = (-1/c)(k + l)F(k, l).
(5.188)
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