The differential operator D2 +12D+ 40 has the form D -2aD+a? + B2 where a = - and B= 2 Therefore D2 + 12D+40 should annihilate the function f = e-6z cos(2a) and g=e-6z sin(2x). We will check that for e-6z cos(2x). Note that when we compute the derivative and second derivative we will get terms that have e-6z sin(2x) in them, so we will have to account for them below. Compute D2 (e-6z cos(2a)), 12D(e 6z cos(2a), and 40e 6z cos(2r). Place the coefficients from the terms e 6z cos(2r) and e-6z sin(2x) in the table. The columns of the table should add to zero. cos(2x) 6z sin(2x) 40 40 12D -12 4. D2 -28 -4
The differential operator D2 +12D+ 40 has the form D -2aD+a? + B2 where a = - and B= 2 Therefore D2 + 12D+40 should annihilate the function f = e-6z cos(2a) and g=e-6z sin(2x). We will check that for e-6z cos(2x). Note that when we compute the derivative and second derivative we will get terms that have e-6z sin(2x) in them, so we will have to account for them below. Compute D2 (e-6z cos(2a)), 12D(e 6z cos(2a), and 40e 6z cos(2r). Place the coefficients from the terms e 6z cos(2r) and e-6z sin(2x) in the table. The columns of the table should add to zero. cos(2x) 6z sin(2x) 40 40 12D -12 4. D2 -28 -4
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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Transcribed Image Text:The differential operator D2 +12D+ 40 has the form D - 2aD+a? + B? where a =
9-
and B= 2
Therefore D2 + 12D+40 should annihilate the function f = e-6z cos(2r) and g=e-6z sin(2x). We will check that for e-6z cos(2x). Note that when
we compute the derivative and second derivative we will get terms that have e-6z sin(2x) in them, so we will have to account for them below.
Compute D2 (e-6z cos(2a)), 12D(e 6z cos(2z), and 40e 6z cos(2r). Place the coefficients from the terms e-6z cos(2r) and e 6z sin(2x) in the table.
The columns of the table should add to zero.
-6z
cos(2x)
-6z sin(2x)
40
40
12D
-12
4.
D2
-28
-4
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