As an illustration of the difficulties that may arise in using the method of undetermined coefficients, consider х + Form the complementary solution to the homogeneous equation. 1 e^(2t) help (formulas) x.(t) = c1 + C2 help (matrices) -1 e^(2t) Show that seeking a particular solution of the form x,(t) = e2"ā, where a = is a constant vector, does not work. In fact, if x, had this form, we would arrive at the following contradiction: a1 help (numbers) and • aj help (numbers) Show that seeking a particular solution of the form xp(t) = te2 ā, where a is a a2 constant vector, does not work either. In fact, if x, had this form, we would arrive at the following contradiction: • aj help (numbers) and aj = help (numbers) and a, = help (numbers)
As an illustration of the difficulties that may arise in using the method of undetermined coefficients, consider х + Form the complementary solution to the homogeneous equation. 1 e^(2t) help (formulas) x.(t) = c1 + C2 help (matrices) -1 e^(2t) Show that seeking a particular solution of the form x,(t) = e2"ā, where a = is a constant vector, does not work. In fact, if x, had this form, we would arrive at the following contradiction: a1 help (numbers) and • aj help (numbers) Show that seeking a particular solution of the form xp(t) = te2 ā, where a is a a2 constant vector, does not work either. In fact, if x, had this form, we would arrive at the following contradiction: • aj help (numbers) and aj = help (numbers) and a, = help (numbers)
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:As an illustration of the difficulties that may arise in using the method of undetermined
coefficients, consider
x +
Form the complementary solution to the homogeneous equation.
1
e^(2t)
help (formulas)
help (matrices)
*(1) =
= c1
-1
e^(2t)
Show that seeking a particular solution of the form x,(t) = e2 ā, where a =
a2
is a
constant vector, does not work. In fact, if x, had this form, we would arrive at the following
d.
contradiction:
help (numbers)
= n
and
help (numbers)
Show that seeking a particular solution of the form xp(t) = te2 a, where a
is a
a2
constant vector, does not work either. In fact, if x, had this form, we would arrive at the
d.
following contradiction:
a, =
a1
help (numbers)
and
aj =
help (numbers)
and
help (numbers)
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