Advanced Engineering Mathematics
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ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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![Compica FOOLS
If the roots to the characteristic equation (3.2) are complex,' then, since the coef-
ficients of the characteristic polynomial are real, the roots are complex conjugates.
They have the form λ = a + ib and =a-ib. The corresponding solutions are
z(t) = e(+) and Z(t)=ea-ib. Using Euler's formula, these solutions become
z(t) = (a+b) =
Z(t)=ea-ib
et [cos(br) + i sin(br)]. and
=e" [cos(bt) - i sin(br)].
where
Since (t) = e-22(1), the solutions are not constant multiples of each other, so
they are linearly independent. Hence using Theorem 1.23, we see that the general
solution is
y(t) = C₁z(t) + C₂Z(r).
(3.8)
The solutions and are complex valued. Such solutions are often preferred (for
2
example, in circuit analysis). However, we are aiming for real valued solutions.
Notice from (3.7) that z and Z are complex conjugates. Written in terms of their
real and imaginary parts, we have
2(t)= y(t) +¡y₂(r) and (r) = y(t)-iys(t).
Thus we have
(3.7)
y(t) = Rez(t) = e cos(br) and y(t) = Imz(t) = esin(bt). (3.10)
+3 Linear, Homogeneous Equations with Constant Coefficienta
using complex (or real) exponentials:
sin z, cos 2, tan r, sinh r, cosh 2, tanh r
(3.9)
3) == (20)+2()) and 32(0)= (200)-2010).
Therefore, by Proposition 1.18. y() and y₂(1), the real and imaginary parts of
z(t), are solutions. Furthermore, these are real valued solutions. Since y(t) =
tan(br) y (r) they are not constant multiples of cach other, so they are linearly
independent. Hence by Theorem 1.23, they form a fundamental set of solutions,
and the general solution can be written as
ya) = Ay (1) + A₂y2(1) Ae" cos(ht) + Age sin(br),
where A, and Ay are arbitrary constants.
We summarize our discussion in the following proposition.
Then, differentiate all of them (in their exponential forms) and verify the standard
derivative formulae.
153](https://content.bartleby.com/qna-images/question/5e343f85-ac09-4dac-ae1f-fa90b444949b/1a8321d3-bd84-490a-bb24-2a1fc96c50ac/fsk6upy_processed.jpeg)
Transcribed Image Text:Compica FOOLS
If the roots to the characteristic equation (3.2) are complex,' then, since the coef-
ficients of the characteristic polynomial are real, the roots are complex conjugates.
They have the form λ = a + ib and =a-ib. The corresponding solutions are
z(t) = e(+) and Z(t)=ea-ib. Using Euler's formula, these solutions become
z(t) = (a+b) =
Z(t)=ea-ib
et [cos(br) + i sin(br)]. and
=e" [cos(bt) - i sin(br)].
where
Since (t) = e-22(1), the solutions are not constant multiples of each other, so
they are linearly independent. Hence using Theorem 1.23, we see that the general
solution is
y(t) = C₁z(t) + C₂Z(r).
(3.8)
The solutions and are complex valued. Such solutions are often preferred (for
2
example, in circuit analysis). However, we are aiming for real valued solutions.
Notice from (3.7) that z and Z are complex conjugates. Written in terms of their
real and imaginary parts, we have
2(t)= y(t) +¡y₂(r) and (r) = y(t)-iys(t).
Thus we have
(3.7)
y(t) = Rez(t) = e cos(br) and y(t) = Imz(t) = esin(bt). (3.10)
+3 Linear, Homogeneous Equations with Constant Coefficienta
using complex (or real) exponentials:
sin z, cos 2, tan r, sinh r, cosh 2, tanh r
(3.9)
3) == (20)+2()) and 32(0)= (200)-2010).
Therefore, by Proposition 1.18. y() and y₂(1), the real and imaginary parts of
z(t), are solutions. Furthermore, these are real valued solutions. Since y(t) =
tan(br) y (r) they are not constant multiples of cach other, so they are linearly
independent. Hence by Theorem 1.23, they form a fundamental set of solutions,
and the general solution can be written as
ya) = Ay (1) + A₂y2(1) Ae" cos(ht) + Age sin(br),
where A, and Ay are arbitrary constants.
We summarize our discussion in the following proposition.
Then, differentiate all of them (in their exponential forms) and verify the standard
derivative formulae.
153
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