APM3706 Assignment 02 Due May 2026 |Ordinary Differential Equations|

UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology


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APM3706: Ordinary Differential Equations

Assignment 02 — 2026

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APM3706
Module Code:
Ordinary Differential Equations
Module Name:
Assignment 02
Assignment:
May 2026
Due Date:




Submitted in partial fulfilment of the requirements for APM3706 — UNISA 2026

,UNISA | APM3706 Assignment 02 – Ordinary Differential Equations



Question 1


(i) Exercise 2.4 (b) from Study Guide


Question: Solve the differential equation:


y ′′ + 4y = 0



Solution:

Assume a solution of the form y = ert . Then y ′′ = r2 ert .

Substituting into the equation:
r2 ert + 4ert = 0


Implying that:
ert (r2 + 4) = 0


Since ert ̸= 0 for all t, the characteristic equation is:


r2 + 4 = 0



Solving:
r2 = −4 =⇒ r = ±2i


Therefore the roots are purely imaginary: r = 0 ± 2i, with α = 0 and β = 2.

The general solution for complex conjugate roots α ± βi is:


y = eαt (c1 cos(βt) + c2 sin(βt))



Therefore the general solution is:


y = c1 cos(2t) + c2 sin(2t)



where c1 and c2 are arbitrary constants.



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,UNISA | APM3706 Assignment 02 – Ordinary Differential Equations



(ii) Exercise 2.32 (c) from Study Guide


Question: Use variation of parameters to solve:

2
y ′′ − y =
1 + et


Solution:


Step 1: Solve the homogeneous equation


The homogeneous equation is:
y ′′ − y = 0


The characteristic equation is:


r2 − 1 = 0 =⇒ (r − 1)(r + 1) = 0 =⇒ r = 1, r = −1



Therefore, the complementary solution is:


yc = c1 et + c2 e−t



So the fundamental set of solutions is:


y1 = et , y2 = e−t



Step 2: Compute the Wronskian


et e−t
W (y1 , y2 ) = = et (−e−t ) − e−t (et ) = −1 − 1 = −2
et −e−t



Step 3: Apply the variation of parameters formulas


The particular solution has the form yp = u1 y1 + u2 y2 , where:




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,UNISA | APM3706 Assignment 02 – Ordinary Differential Equations




−y2 g(t) y1 g(t)
u′1 = , u′2 =
W W

2
with g(t) = .
1 + et
Computing u′1 :
2
−e−t · −t
u′1 = 1 + et = e
−2 1 + et

Multiply numerator and denominator by et :

1 t e
t e−t · et 1
u′1 = t 2t
· e · t
= t t
= t
e +e e (1 + e ) · e e (1 + et )


1
Rewrite using partial fractions. Note that e−t = , so:
et

e−t
u′1 =
1 + et


Integrating:
e−t
Z
u1 = dt
1 + et

1 1+v
Let v = e−t , so dv = −e−t dt, and et = , giving 1 + et = ... Using a cleaner substitution,
v v
let s = et , ds = et dt = s dt, so dt = ds/s:

s−1 ds
Z Z
1
u1 = · = ds
1+s s s2 (1+ s)


1 A B C
Using partial fractions: = + 2+ .
s2 (1 + s) s s 1+s
Multiplying through by s2 (1 + s):


1 = As(1 + s) + B(1 + s) + Cs2



Setting s = 0: 1 = B, so B = 1.

Setting s = −1: 1 = C, so C = 1.

Expanding and matching s2 coefficient: 0 = A + C, so A = −1.



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,UNISA | APM3706 Assignment 02 – Ordinary Differential Equations


Therefore:
Z  
−1 1 1 1
u1 = + 2+ ds = − ln s − + ln(1 + s) + K
s s 1+s s

Substituting back s = et :
u1 = −t − e−t + ln(1 + et )


Computing u′2 :
2
et · t
u′2 = 1 + et = −e
−2 1 + et

Integrating (let s = 1 + et , ds = et dt):

−et
Z
u2 = dt = − ln(1 + et )
1 + et


Step 4: Form the particular solution


yp = u1 y1 + u2 y2

yp = −t − e−t + ln(1 + et ) et + − ln(1 + et ) e−t





yp = −tet − 1 + et ln(1 + et ) − e−t ln(1 + et )

yp = −tet − 1 + (et − e−t ) ln(1 + et )


The −1 is absorbed into the complementary solution constants, so the general solution is:



y = c1 et + c2 e−t − tet + (et − e−t ) ln(1 + et )




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, UNISA | APM3706 Assignment 02 – Ordinary Differential Equations



(iii) Exercise 3.16 from Study Guide


Question: Solve the system using the eigenvalue method:
 
1 1
Ẋ =  X
4 1


Solution:


Step 1: Find the eigenvalues


det(A − λI) = 0
 
1−λ 1
det   = (1 − λ)2 − 4 = 0
4 1−λ

1 − 2λ + λ2 − 4 = 0 =⇒ λ2 − 2λ − 3 = 0 =⇒ (λ − 3)(λ + 1) = 0


Therefore λ1 = 3 and λ2 = −1.


Step 2: Find the eigenvectors


For λ1 = 3:

  
−2 1 k
(A − 3I)k = 0 =⇒    1 = 0
4 −2 k2

From the first row: −2k1 + k2 = 0 =⇒ k2 = 2k1 .
 
1
Setting k1 = 1: k1 =  .
2

For λ2 = −1:

  
2 1 k
(A + I)k = 0 =⇒    1 = 0
4 2 k2

From the first row: 2k1 + k2 = 0 =⇒ k2 = −2k1 .




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