Chapter 4 Even-Numbered Homework Solutions
4.1
16.
dy
d2 y
+4
+ 20y = e−t/2
dt2
dt
dy
d2 y
+4
+ 20y = 0. First compute the general solution for yh (t).
dt2
dt
2
The characteristic equation is s + 4s + 20 = 0, which gives s = −2 ± 4i. So, yh (t) = e−2t (k1 cos 4t + k2 sin 4t).
(a) The associated homogeneous equation is
Now, let yp (t) be a particular solution for αe−t/2 . Then we have:
d2
d
(αe−t/2 ) + 4 (αe−t/2 ) + 20(αe−t/2 ) = e−t/2
dt2
dt
−1 −t/2
1
⇒ αe−t/2 + 4
αe
+ 20(αe−t/2 ) = e−t/2
4
2
α
⇒ e−t/2
− 2α + 20α = e−t/2
4
α
⇒ − 2α + 20α = 1
4
4
.
⇒α=
73
4 −t/2
e
. Since y(t) = yh (t) + yp (t), we now know that:
73
4
y(t) = e−2t (k1 cos 4t + k2 sin 4t) + e−t/2 .
73
4
−4
(b) Let t = 0. Then y(0) = 0 = k1 + , which implies that k1 =
. Calculate k2 using y 0 (0) = 0:
73
73
2
y 0 (0) = 0 = −2k1 + 4k2 −
73
−4
2
⇒ −4k2 = −2
−
73
73
2
6
8
−
=
⇒ −4k2 =
73 73
73
3
⇒ k2 = −
.
146
−4
3
4
−2t
The solution satisfying the initial conditions is y(t) = e
cos 4t −
sin 4t + e−t/2 .
73
146
73
So, yp (t) =
(c) As t → ∞, y(t) → 0.
38.
d2 y
dy
+3
+ 2y = e−t − 4
2
dt
dt
(a) The associated homogeneous equation is
d2 y
dy
+ 3 + 2y = 0, whose characteristic equation is s2 + 3s + 2y = 0,
2
dt
dt
1
which gives s = −1, −2. Thus yh (t) = k1 e−t + k2 e−2t . Let yp (t) = αte−t + c be a particular solution. Then:
d2
d
(αte−t + c) + 3 (αte−t + c) + 2(αte−t + c) = e−t − 4
dt2
dt
You should get α = 1 and c = −2. Thus, y(t) = k1 e−t + k2 e−2t + te−t − 2.
(b) Using the initial conditions we get:
k1 + k2 − 2 = 0 and −k1 − 2k2 + 1 = 0.
Thus, k1 = 3 and k2 = −1. So, y(t) = 3e−t − e−2t + te−t − 2.
(c) As t → ∞, y(t) → −2.
4.2
6. Find the general solution of the given equation:
dy
d2 y
+6
+ 8y = −4 cos 3t
dt2
dt
d2 y
dy
+6
+ 8y = 0. The characteristic equation is s2 + 6s + 8 = 0,
dt2
dt
which means that s = −4, −2. Thus, yh (t) = k1 e−2t + k2 e−4t . Now, let yp (t) = a cos 3t + b sin 3t. Then we
have:
d2
d
(a cos 3t + b sin 3t) + 4 (a cos 3t + b sin 3t) + 20(a cos 3t + b sin 3t) = −4 cos 3t
dt2
dt
Its associated homogeneous linear equation is
⇒ −9a cos 3t − 9b sin 3t + 6(−3a sin 3t + 3b cos 3t) + 8(a cos 3t + b sin 3t) = −4 cos 3t
(−a + 18b) cos 3t + (−18a − b) sin 3t = −4 cos 3t.
−72
4
and b =
. Since y(t) = yh (t) + yp (t), the general solution is:
325
325
4
72
y(t) = k1 e−2t + k2 e−4t +
cos 3t −
sin 3t.
325
325
So, −a + 18b = 4 and −18a − b = 0 ⇒ a =
12. Find the solution of the given initial-value problem:
d2 y
dy
+6
+ 8y = 2 cos 3t,
dt2
dt
y(0) = y 0 (0) = 0.
We know from 4.2.6 above that yh (t) = k1 e−2t + k2 e−4t . Let yp (t) = a cos 3t + b sin 3t. We have:
−a + 18b = 2 and −18a − b = 0
36
2
and b =
.
325
325
2
36
Thus, our general solution is y(t) = k1 e−2t + k2 e−4t −
cos 3t +
sin 3t. Using the initial condition y(0) = 0,
325
325
we get:
⇒a=−
0 = k1 + k2 −
Solving these equations gives k1 =
2
108
and 0 = −2k1 − 4k2 +
.
325
325
−2
4
and k2 =
.
13
25
2
4.3
10. Compute the solution of the given initial-value problem:
d2 y
+ 4y = 3 cos(2t),
y(0) = y 0 (0) = 0.
dt2
d2 y
+ 4y = 3e2it . Guess yc (t) = ate2it as a particular solution.
Consider the complex version of this equation:
dt2
(Why won’t yc (t) = ae2it work? Check it!) Then yc00 (t) = ae2it (4i − 4t). Plugging this back into the equation we get:
ae2it (4i − 4t) + 4ate2it = 3e2it
4iae2it = 3e2it
3
−3i
−3i
=
. Taking the real part of yc (t) =
it(cos(2t) + i sin(2t)), we obtain
So, yc (t) is a solution if a =
4i
4
4
3
y(t) = t sin(2t) as a solution of the original equation. To find the general solution of the homogeneous equation,
4
we note that the characteristic polynomial is s2 + 4 = 0, which has roots s = ±2i. Hence, the general solution of
the original equation is:
3
y(t) = k1 cos(2t) + k2 sin(2t) + t sin(2t)
4
Using the initial conditions y(0) = y 0 (0) = 0, find that k1 = 0 and k2 = 0, which means the solution of the IVP is:
y(t) =
3
t sin(2t).
4
16. Use the following equation:
d2 y
+ 11y = 2 cos(3t)
dt2
√
(a) The characteristic
polynomial of the unforced equation is s2 + 11, which has roots
√
√ s = ±i 11. So the natural
11
11 − 3
3
frequency is
and the forcing frequency is
. The frequency of the beats is
.
2π
2π
4π
√
11 + 3
.
(b) The frequency of the rapid oscillations is
4π
3