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Math 524 - Elementary Differential Equations
Instructor: J. Metcalfe
Due: March 27, 2008
Assignment 17
Section 5.4, # 8 Find the general solution to
x
′
(^) x.
Since
0 = det A − λI = −(λ − 13)
2 (λ − 7),
we have that λ = 7 and λ = 13 are the eigenvalues. The latter has multiplicity 2.
Since
A − 7 I ∼
we have that
ξ =
is an eigenvector.
Since
A − 13 I ∼
we see that
ξ 1 =
(^) , ξ 2 =
are two linearly independent eigenvectors.
Thus, the general solution is
x
′ = c 1
(^) e^7 t^ + c 2
(^) e^13 t^ + c 3
(^) e^13 t.
Section 5.4, #20 Find the general solution to
x
′
x.
Since the matrix is triangular, it is clear that λ = 2 is the only eigenvalue and it has
multiplicity 4. And,
A − 2 I ∼
Thus, the defect is 3. That is, ξ = e 1 is the only independent eigenvector we can find.
We want to find the generalized eigenvectors that we can use to construct the re-
maining independent solutions. We first solve (A − 2 I)η = ξ. Using an augmented
matrix, we see easily that we may take
η =
We next solve (A − 2 I)μ = η. Here, we take μ = e 3. Finally, we solve (A − 2 I)ν = μ.
Here, we take
ν =
We may now form the general solution
x(t) = c 1
e
2 t
te
2 t
e
2 t
t
2
e
2 t
te
2 t
e
2 t
t^3
e
2 t
t^2
e
2 t
te
2 t
e
2 t
Section 5.4, #22 Find the general solution to
x
′
x.
Here, we have
0 = det(A − λI) = (1 − λ)
4 .
