Chapter 4 Even-Numbered Homework Solutions

4.1

16.

d2y

dt2+ 4

dy

dt+ 20y = e−t/2

(a) The associated homogeneous equation isd2y

dt2+ 4

dy

dt+ 20y = 0. First compute the general solution for yh(t).

The characteristic equation is s2 + 4s+ 20 = 0, which gives s = −2± 4i. So, yh(t) = e−2t(k1 cos 4t+ k2 sin 4t).

Now, let yp(t) be a particular solution for αe−t/2. Then we have:

d2

dt2(αe−t/2) + 4

d

dt(αe−t/2) + 20(αe−t/2) = e−t/2

⇒ 1

4αe−t/2 + 4

(−1

2αe−t/2

)+ 20(αe−t/2) = e−t/2

⇒ e−t/2(α

4− 2α+ 20α

)= e−t/2

⇒ α

4− 2α+ 20α = 1

⇒ α =4

73.

So, yp(t) =4

73e−t/2. Since y(t) = yh(t) + yp(t), we now know that:

y(t) = e−2t(k1 cos 4t+ k2 sin 4t) +4

73e−t/2.

(b) Let t = 0. Then y(0) = 0 = k1 +4

73, which implies that k1 =

−4

73. Calculate k2 using y′(0) = 0:

y′(0) = 0 = −2k1 + 4k2 −2

73

⇒ −4k2 = −2

(−4

73

)− 2

73

⇒ −4k2 =8

73− 2

73=

6

73

⇒ k2 = − 3

146.

The solution satisfying the initial conditions is y(t) = e−2t(−4

73cos 4t− 3

146sin 4t

)+

4

73e−t/2.

(c) As t→∞, y(t)→ 0.

38.

d2y

dt2+ 3

dy

dt+ 2y = e−t − 4

(a) The associated homogeneous equation isd2y

dt2+ 3

dy

dt+ 2y = 0, whose characteristic equation is s2 + 3s+ 2y = 0,

1

which gives s = −1,−2. Thus yh(t) = k1e−t + k2e

−2t. Let yp(t) = αte−t + c be a particular solution. Then:

d2

dt2(αte−t + c) + 3

d

dt(αte−t + c) + 2(αte−t + c) = e−t − 4

You should get α = 1 and c = −2. Thus, y(t) = k1e−t + k2e

−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:

d2y

dt2+ 6

dy

dt+ 8y = −4 cos 3t

Its associated homogeneous linear equation isd2y

dt2+ 6

dy

dt+ 8y = 0. The characteristic equation is s2 + 6s+ 8 = 0,

which means that s = −4,−2. Thus, yh(t) = k1e−2t + k2e

−4t. Now, let yp(t) = a cos 3t+ b sin 3t. Then wehave:

d2

dt2(a cos 3t+ b sin 3t) + 4

d

dt(a cos 3t+ b sin 3t) + 20(a cos 3t+ b sin 3t) = −4 cos 3t

⇒ −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.

So, −a+ 18b = 4 and −18a− b = 0⇒ a =4

325and b =

−72

325. Since y(t) = yh(t) + yp(t), the general solution is:

y(t) = k1e−2t + k2e

−4t +4

325cos 3t− 72

325sin 3t.

12. Find the solution of the given initial-value problem:

d2y

dt2+ 6

dy

dt+ 8y = 2 cos 3t, y(0) = y′(0) = 0.

We know from 4.2.6 above that yh(t) = k1e−2t + k2e

−4t. Let yp(t) = a cos 3t+ b sin 3t. We have:

−a+ 18b = 2 and −18a− b = 0

⇒ a = − 2

325and b =

36

325.

Thus, our general solution is y(t) = k1e−2t + k2e

−4t − 2

325cos 3t+

36

325sin 3t. Using the initial condition y(0) = 0,

we get:

0 = k1 + k2 −2

325and 0 = −2k1 − 4k2 +

108

325.

Solving these equations gives k1 =−2

13and k2 =

4

25.

2

4.3

10. Compute the solution of the given initial-value problem:d2y

dt2+ 4y = 3 cos(2t), y(0) = y′(0) = 0.

Consider the complex version of this equation:d2y

dt2+ 4y = 3e2it. Guess yc(t) = ate2it as a particular solution.

(Why won’t yc(t) = ae2it work? Check it!) Then y′′c (t) = ae2it(4i−4t). Plugging this back into the equation we get:

ae2it(4i− 4t) + 4ate2it = 3e2it

4iae2it = 3e2it

So, yc(t) is a solution if a =3

4i=−3i

4. Taking the real part of yc(t) =

−3i

4it(cos(2t) + i sin(2t)), we obtain

y(t) =3

4t sin(2t) as a solution of the original equation. To find the general solution of the homogeneous equation,

we note that the characteristic polynomial is s2 + 4 = 0, which has roots s = ±2i. Hence, the general solution ofthe original equation is:

y(t) = k1 cos(2t) + k2 sin(2t) +3

4t sin(2t)

Using the initial conditions y(0) = y′(0) = 0, find that k1 = 0 and k2 = 0, which means the solution of the IVP is:

y(t) =3

4t sin(2t).

16. Use the following equation:

d2y

dt2+ 11y = 2 cos(3t)

(a) The characteristic polynomial of the unforced equation is s2 + 11, which has roots s = ±i√

11. So the natural

frequency is

√11

2πand the forcing frequency is

3

2π. The frequency of the beats is

√11− 3

4π.

(b) The frequency of the rapid oscillations is

√11 + 3

4π.

3