MAT3700 Assignment 2 Semester 1 2026 - Due May 2026

MAT3700
ASSIGNMENT 2 2026
DUE MAY JUNE 2026

,MAT3700 ASSESSMENT 2 2026

DUE MAY/JUNE 2026




QUESTION 1

Find the general solutions of the following differential equations using D-operator
methods:

1.1 (D 2 − 4D + 4)y = e2x + e−2x

1.2 (D 2 + 2D + 2)y = e−x sin(2x) ​




1.3 (D 2 + 6D + 10)y = x 2 + sin(3x)




QUESTION 1 (D-operator methods)
​ ​




1.1 (D 2 − 4D + 4)y = e2x + e−2x


Characteristic equation: m2 − 4m + 4 = 0 ⇒ (m − 2)2 = 0 → repeated root m = 2.
Complementary function:

yc = (A + Bx)e2x
​




Particular integral:
2
For e2x : use (D−2)
1
2 e = e2x ⋅ D12 ⋅ 1 = e2x ⋅ x2 .
2x
​ ​




For e−2x : substitute D = −2: (−2)2 − 4(−2) + 4 = 4 + 8 + 4 = 16, so 1 −2x
16
​ e .
Thus

x2 2x 1 −2x
yp = e + e
2 16
​




General solution:
​




x2 1
y = e (A + Bx + ) + e−2x
2x
2 16

,1.2 (D 2 + 2D + 2)y = e−x sin(2x)


Characteristic: m2 + 2m + 2 = 0 ⇒ m = −1 ± i

yc = e−x (C cos x + D sin x)

For PI, use shift theorem:
e−x sin(2x): P (D) = D2
+ 2D + 2, replace D by D − 1:
P (D − 1) = (D − 1) + 2(D − 1) + 2 = D2 + 1.
2
Thus

1 1 sin 2x 1
e−x sin(2x) = e−x 2 sin(2x) = e−x ( ) = − e−x sin(2x)
P (D) D +1 −4 + 1 3

So

1
yp = − e−x sin(2x)
3



General solution:

1
y = e−x (C cos x + D sin x) − e−x sin(2x)
3

, 1.3 (D 2 + 6D + 10)y = x2 + sin(3x)




Characteristic: m2 + 6m + 10 = 0 ⇒ m = −3 ± i

yc = e−3x (E cos x + F sin x)
​




For polynomial part: let yp1 = ax2 + bx + c. Substitute:
​




D2 (ax2 + bx + c) = 2a, D(ax2 + bx + c) = 2ax + b. Then
(2a) + 6(2ax + b) + 10(ax2 + bx + c) = 10ax2 + (12a + 10b)x + (2a + 6b + 10c) = x2 + 0x +
0
.
Compare: 10a = 1 ⇒ a = 0.1; 12a + 10b = 0 ⇒ 1.2 + 10b = 0 ⇒ b = −0.12;
2a + 6b + 10c = 0 ⇒ 0.2 − 0.72 + 10c = 0 ⇒ c = 0.052.
So



yp1 = 0.1x2 − 0.12x + 0.052
​




For sin(3x): substitute D 2
= −9:
​ ​




1 1
−9+6D+10
sin(3x) = 1+6D sin(3x). Multiply numerator and denominator by 1 − 6D:
​ ​




1−6D 1−6D 1
= 1−36D 2 sin(3x) = 1+324 sin(3x) = 325 (sin 3x − 6 ⋅ 3 cos 3x).
​




Thus ​




1
yp2 = (sin 3x − 18 cos 3x)
325
​




General solution:

1
y = e−3x (E cos x + F sin x) + 0.1x2 − 0.12x + 0.052 + (sin 3x − 18 cos 3x)
325