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306 CHAPTER 10 Systems of Linear Differential Equations

Then
−1
3/2 −1 150
C =
1 1 50

2/5 2/5 150 80
= = .
−2/5 3/5 50 −30
The solution of the initial value problem is
−t/10 −3t/5
(3/2)e −e 80
X(t) = −t/10 −3t/5
e e −30
−t/10 −3t/5
120e + 30e
= −t/10 −3t/5 .
80e − 30e
As t →∞, x 1 (t) → 0 and x 2 (t) → 0, as we might expect.

10.2.1 Solution When A Has a Complex Eigenvalue
We used Euler’s formula to write real-valued solutions of the second-order linear homogeneous
constant coefficient differential equation when the characteristic equation has complex roots. We
will follow a similar procedure for systems when the matrix of coefficients has (at least some)
complex eigenvalues.
Since A is assumed to have real elements, the characteristic polynomial has real coefficients,
so complex roots must occur in complex conjugate pairs. If λ is a complex eigenvalue with eigen-
λt
vector ξ, then λ is also an eigenvalue, with complex eigenvector ξ. Therefore, ξe and ξe λt
are solutions. By taking linear combinations of any two such solutions, we obtain the
following.

THEOREM 10.7 Solutions When Complex Eigenvalues Occur

Let A be an n ×n matrix of real numbers. Let α +iβ be a complex eigenvalue with corresponding
eigenvector U + iV, in which U and V are real n × 1 matrices. Then
αt
e [cos(βt)U − sin(βt)V]
and
αt
e [sin(βt)U + cos(βt)V]
are linearly independent solutions of X = AX.

Proof We know that α −iβ is also an eigenvalue with eigenvector U−iV. Write two solutions:
1 (t) = e (α+iβ)t (U + iV)
αt
= e (cos(βt) + i sin(βt))(U + iV)
αt
αt
= e (cos(βt)U − sin(βt)V) + ie (sin(βt)U + cos(βt)V)
and
2 (t) = e (α−iβ)t (U − iV)
αt
= e (cos(βt) − i sin(βt))(U − iV)
αt
αt
= e (cos(βt)U − sin(βt)V) + ie (−cos(βt)V − sin(βt)U).

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October 14, 2010 20:32 THM/NEIL Page-306 27410_10_ch10_p295-342