Download Week 16 Homework

Mathematics for Engineers and Scientists (MATH1551)
75. Find the general solution of each of the following.
a) y 00 + 2y 0 − 15y = 0,
b) 2y 00 + 3y 0 − 2y = 0,
c) y 00 − 6y 0 + 25y = 0
d) y 00 + 6y 0 + 9y = 0,
e) y 00 + y 0 − 2y = 2e−x ,
f) y 00 + y 0 − 2y = ex ,
g) y 00 + y 0 − 6y = 52 cos 2x,
h) 2y 00 + 3y 0 − 2y = sin x,
i) y 00 + y = 2 sin x,
j) y 00 − 2y 0 + 2y = ex cos x.
Solution:
(e) The auxiliary equation λ2 + λ − 2 = (λ − 1)(λ + 2) = 0 has real roots λ = 1, −2.
Thus the complementary function is yc = Aex + Be−2x where A, B are constants.
Now look for a particular solution of the form yp = ae−x . Then
yp00 + yp0 − 2yp = ae−x − ae−x − 2ae−x
= −2ae−x = 2e−x
Equating coefficients gives a = −1 and thus the general solution is
y = Aex + Be−2x − e−x .
77. Solve 2y 00 + 5y 0 + 2y = 0 with initial conditions y(0) = 4, y 0 (0) = −1/2.
Solution:
The auxiliary equation 2λ2 + 5λ + 2 = (2λ + 1)(λ + 2) = 0 has real roots λ = −1/2, −2.
Thus the general solution is y = Ae−x/2 + Be−2x where A, B are constants.
Now fit the initial conditions: as y 0 = − 12 Ae−x/2 − 2Be−2x we have
y(0) = A + B = 4
1
1
y 0 (0) = − A − 2B = −
2
2
The required solution is:
=⇒
A = 5 and B = −1.
y = 5e−x/2 − e−2x .
81. Solve y 00 − 4y 0 + 29y = 0 with initial conditions y(0) = 2, y 0 (0) = −1.
Solution:
The auxiliary equation λ2 − 4λ + 29 = 0 has complex roots λ =
√
8± 42 −4·29
2
= 2 ± 5i.
Thus the general solution is y = e2x (A cos 5x + B sin 5x) where A, B are constants.
1
Now fit the initial conditions: as
y 0 = 2e2x (A cos 5x + B sin 5x) + e2x (−5A sin 5x + 5B cos 5x)
we have
y(0) = A = 2
y 0 (0) = 2A + 5B = −1
The required solution is:
=⇒
A = 2 and B = −1.
y = e2x (2 cos 5x − sin 5x).
85. Solve y 00 + 6y 0 + 9y = 0 with initial conditions y(0) = 2, y 0 (0) = −5.
Solution:
The auxiliary equation λ2 + 6λ + 9 = (λ + 3)2 = 0 has equal roots λ = −3, −3.
Thus the general solution is y = e−3x (Ax + B) where A, B are constants.
Now fit the initial conditions: as y 0 = −3e−3x (Ax + B) + Ae−3x we have
y(0) = B = 2
y 0 (0) = −3B + A = −5
The required solution is:
=⇒
y = e−3x (x + 2).
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A = 1 and B = 2.