Mathematics for Engineers and Scientists (MATH1551)
92. Solve y 00 − 6y 0 + 10y = 6 sin 2x + 18 cos 2x, with y(0) = 2, y 0 (0) = 3.
Solution:
The auxiliary equation λ2 − 6λ + 10 = 0 has complex roots λ =
√
6± 36−40
2
= 3 ± i.
Thus the complementary function is yc = e3x (A cos x + B sin x) where A, B are constants.
Now look for a particular solution of the form yp = a cos 2x + b sin 2x. Then
yp00 + 6yp0 + 10yp = (−4a cos 2x − 4b sin 2x) − 6(−2a sin 2x + 2b cos 2x) + 10(a cos 2x + b sin 2x)
= (6a − 12b) cos 2x + (12a + 6b) sin 2x
= 18 cos 2x + 6 sin 2x.
Equating coefficients: 6a − 12b = 18 and 12a + 6b = 6, so a = 1, b = −1.
Thus the general solution is y = e3x (A cos x + B sin x) + cos 2x − sin 2x.
Now fit the initial conditions: y(0) = A + 1 = 2, so A = 1. Also,
y 0 = 3e3x (A cos x + B sin x) + e3x (−A sin x + B cos x) − 2 sin 2x − 2 cos 2x
so y 0 (0) = 3A + B − 2 = 3, hence B = 2.
The required solution is
y = e3x (cos x + 2 sin x) + cos 2x − sin 2x.
95. Solve y 00 + 2y 0 + y = x2 + 4x + 1, y(0) = 0, y 0 (0) = 1.
Solution:
The auxiliary equation λ2 + 2λ + 1 = 0 has equal roots λ = −1, −1.
Thus the complementary function is yc = e−x (Ax + B) where A, B are constants.
Now look for a particular solution of the form yp = ax2 + bx + c. Then
yp00 + 2yp0 + yp = (2a) + 2(2ax + b) + (ax2 + bx + c)
= ax2 + (4a + b)x + (2a + 2b + c)
= x2 + 4x + 1.
Equating coefficients: a = 1, 4a + b = 4, and 2a + 2b + c = 1 so a = 1, b = 0 and c = −1.
Thus the general solution is y = e−x (Ax + B) + x2 − 1.
Now fit the initial conditions: y(0) = B − 1 = 0, so B = 1. Also,
y 0 = −e−x (Ax + B) + Ae−x + 2x
so y 0 (0) = −B + A = 1, hence A = 2.
The required solution is
y = e−x (2x + 1) + x2 − 1.
1
103. Use Gaussian Elimination to find the inverses of the following matrices:






4
3
−6
3
1
−1
3
3
−1
a)
b)
c)
1 −1
A= 1
A = −4 1 2 
A = −4 −3 2 
−2 −2 3
−2 0 1
−2 −2 1


7 −1 3
A = −4 1 −2
8
0
3
d)
e)


5
5 −2
A = −4 −3 2 
−2 −2 1
Solution:
Note: for each of these, there are many possible choices for row operations.
a)




4
3 −6 1 0 0
1
1 −1 0 1 0
R ↔R2
1
4
1 −1 0 1 0 −−1−−→
3 −6 1 0 0
−2 −2 3 0 0 1
−2 −2 3 0 0 1



1 1 −1 0 1 0
1 1 −1 0 1 0
R →−R2
R2 →R2 −4R1
0 1 2 −1 4 0
−−
−−−−−→ 0 −1 −2 1 −4 0 −−2−−−→
R3 →R3 +2R1
0 2 1
0 0
1 0 2 1
0 0 1





1 1 0 0 3 1
1 0 0 1 3 3
R2 →R2 −2R3
R →R −R2
0 1 0 −1 0 −2
−−
−−−−−→ 0 1 0 −1 0 −2 −−1−−−1−−→
R1 →R1 +R3
0 0 1 0 2 1
0 0 1 0 2 1
Hence the inverse is

A−1

1 3 3
= −1 0 −2
0 2 1
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