APM3706 Assignment 3 Complete Solutions UNISA2026 Ordinary Differential Equations

APM3706
ASSIGNMENT 3
Ordinary Differential Equations

FULL
SOLUTIONS
COMPLETE SOLUTIONS

MEMORANDUM
UNISA 2026
Page 1 of 19

,SOLUTIONS

Question 1

(1) Prove Theorem 4.5

Theorem 4.5: Φ is a fundamental matrix of 𝑋̇ = 𝐴𝑋 iff

(a) Φ is a solution of 𝑋̇ = 𝐴𝑋, i.e. Φ̇(𝑡) = 𝐴Φ(𝑡);
(b) det Φ(t₀) ≠ 0.




Page 2 of 19

, Proof:

(⇒) Suppose Φ is a fundamental matrix of 𝑋̇ = 𝐴𝑋. By definition, a fundamental matrix is
a matrix whose columns form a linearly independent set of solutions. Therefore, each
column of Φ is a solution of the system, so Φ̇(𝑡) = 𝐴Φ(𝑡). Also, since the columns are
linearly independent, the determinant of Φ(t) is nonzero for all t in the interval, and in
particular det Φ(t₀) ≠ 0.

(⇐) Suppose (a) and (b) hold. Condition (a) means each column of Φ is a solution of the
system. Condition (b) means that det Φ(t₀) ≠ 0, so the columns are linearly independent
at t₀. By the theory of linear systems, linear independence at one point implies linear
independence on the entire interval. Therefore, the columns form a set of n linearly
independent solutions, so Φ is a fundamental matrix. □

0 1
(2) Verify Theorem 4.5 for the system with 𝐀 = [ ].
−1 −2

First, find the eigenvalues of A. The characteristic equation is ∣ 𝐴 − 𝜆𝐼 ∣= 0:

−𝜆 1
det [ ] = (−𝜆)(−2 − 𝜆) − (1)(−1) = 𝜆2 + 2𝜆 + 1 = (𝜆 + 1)2 = 0.
−1 −2 − 𝜆


So λ = -1 is a repeated eigenvalue. For λ = -1, solve (A - (-1)I)v = 0:

1 1 𝑣1 0
[ ] [𝑣 ] = [ ] ⇒ 𝑣1 + 𝑣2 = 0.
−1 −1 2 0


Choose v = [1; -1]. For a repeated eigenvalue, we need a generalized eigenvector w such
that (A + I)w = v:

1 1 𝑤1 1
[ ] [ ] = [ ] ⇒ 𝑤1 + 𝑤2 = 1.
−1 −1 𝑤2 −1


Choose w = [1; 0]. Then a fundamental matrix is

𝑒 −𝑡 (𝑡 + 1)𝑒 −𝑡
Φ(𝑡) = [ ].
−𝑒 −𝑡 −𝑡𝑒 −𝑡


Check condition (a):
−𝑡
−𝑒 −𝑡𝑒 −𝑡
Φ̇(𝑡) = [ −𝑡 ].
𝑒 (𝑡 − 1)𝑒 −𝑡

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