APM3706 ASSIGNMENT 2 COMPLETE SOLUTIONS UNISA 2026 ORDINARY DIFFERENTIAL EQUATIONS

APM3706
ASSIGNMENT 2
Ordinary Differential Equations

FULL
SOLUTIONS
COMPLETE SOLUTIONS

MEMORANDUM
UNISA 2026
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,SOLUTIONS:
Question 1

(i) Exercise 2.4 (b)

We find the eigenvalues and eigenvectors of

−1 2 2
𝐴=[ 2 2 2 ].
−3 −6 −6



The characteristic polynomial is

−1 − 𝜆 2 2
det⁡(𝐴 − 𝜆𝐼) = det⁡ [ 2 2−𝜆 2 ] = −𝜆(𝜆 + 2)(𝜆 + 3),
−3 −6 −6 − 𝜆



so the eigenvalues are 𝜆 = 0, −2, −3.

For 𝜆 = 0: solve 𝐴𝐯 = 0. From the first two equations, subtract to get −3𝑥 = 0 ⇒ 𝑥 = 0.
Then 2𝑦 + 2𝑧 = 0 ⇒ 𝑦 = −𝑧. Taking 𝑧 = 1 gives eigenvector (0, −1,1)𝑇 .




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, For 𝜆 = −2: solve (𝐴 + 2𝐼)𝐯 = 0. The second equation minus twice the first gives −2𝑧 =
0 ⇒ 𝑧 = 0. Then 𝑥 + 2𝑦 = 0 ⇒ 𝑥 = −2𝑦. Choose 𝑦 = 1: eigenvector (−2,1,0)𝑇 .

For 𝜆 = −3: solve (𝐴 + 3𝐼)𝐯 = 0. The second equation minus the first gives 3𝑦 = 0 ⇒ 𝑦 =
0. Then 𝑥 + 𝑧 = 0 ⇒ 𝑧 = −𝑥. Choose 𝑥 = 1: eigenvector (1,0, −1)𝑇 .

Thus the eigenvectors are (0, −1,1)𝑇 for 𝜆 = 0, (−2,1,0)𝑇 for 𝜆 = −2,
and (1,0, −1)𝑇 for 𝜆 = −3.




(ii) Exercise 2.32 (c)

We solve 𝐱̇ = 𝐴𝐱 with

4 −3 −2 0
𝐴 = [2 −1 −2] , 𝐱(1) = [0].
3 −3 −1 1



The characteristic polynomial is −𝜆3 + 2𝜆2 + 𝜆 − 2 = −(𝜆 − 1)(𝜆 − 2)(𝜆 + 1) = 0, so
eigenvalues 𝜆 = 1,2, −1.

For 𝜆 = 1: solve (𝐴 − 𝐼)𝐯 = 0. The second equation gives 𝑥 − 𝑦 − 𝑧 = 0. Substitute 𝑧 =
𝑥 − 𝑦 into the first: 3𝑥 − 3𝑦 − 2(𝑥 − 𝑦) = 𝑥 − 𝑦 = 0 ⇒ 𝑥 = 𝑦, then 𝑧 = 0. So
eigenvector (1,1,0)𝑇 .

For 𝜆 = 2: solve (𝐴 − 2𝐼)𝐯 = 0. The third equation gives 𝑥 − 𝑦 − 𝑧 = 0 ⇒ 𝑦 = 𝑥 − 𝑧.
Substitute into the first: 2𝑥 − 3(𝑥 − 𝑧) − 2𝑧 = −𝑥 + 𝑧 = 0 ⇒ 𝑧 = 𝑥, then 𝑦 = 0. So
eigenvector (1,0,1)𝑇 .

For 𝜆 = −1: solve (𝐴 + 𝐼)𝐯 = 0. The second equation gives 2𝑥 − 2𝑧 = 0 ⇒ 𝑥 = 𝑧. The
third gives 3𝑥 − 3𝑦 = 0 ⇒ 𝑥 = 𝑦. So 𝑥 = 𝑦 = 𝑧, eigenvector (1,1,1)𝑇 .

The general solution is

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