HW-4 Solution:

S4.3 - 3, 12, 19, 28, 36; S4.4 - 11, 14, 16

S4.3 - Q3

y′′ + 5y′ + 6y = 0. The characteristic equation is

λ2 + 5λ+ 6 = 0 ⇐⇒ (λ+ 2)(λ+ 3) = 0 ⇐⇒ λ = −2 or − 3.

A fundamental set of solutions is given by y1(t) = e−2t and y2(t) = e−3t.

The general solution is y(t) = C1e−2t + C2e

−3t.

S4.3 - Q12

y′′ + 2y′ + 17y = 0. The characteristic equation is

λ2 + 2λ+ 17 = 0 ⇐⇒ λ =−2±

√4− 4 · 1 · 17

2

λ =−2± 8i

2= −1± 4i.

A fundamental set of solutions is given by y1(t) = e−t cos(4t) and y2(t) = e−t sin(4t).

The general solution is y(t) = e−t(C1 cos(4t) + C2 sin(4t)

).

S4.3 - Q19

4y′′ + 4y′ + y = 0. The characteristic equation is

4λ2 + 4λ+ 1 = 0 ⇐⇒ (2λ+ 1)2 = 0

⇐⇒ λ = −1

2(repeated root).

A fundamental set of solutions is given by y1(t) = e−t/2 and y2(t) = te−t/2.

The general solution is y(t) = C1e−t/2 + C2te

−t/2.

S4.3 - Q28

y′′ + 25y = 0, y(0) = 1, y′(0) = −1. The characteristic equation is

λ2 + 25 = 0 ⇐⇒ λ = ±5i.

A fundamental set of solutions is given by y1(t) = cos(5t) and y2(t) = sin(5t).

The general solution is y(t) = C1 cos(5t) + C2 sin(5t).

Put y(0) = 1. 1 = C1 + 0 ⇐⇒ C1 = 1.

Note y′(t) = −5C1 sin(5t) + 5C2 cos(5t) = −5 sin(5t) + 5C2 cos(5t).

Put y′(0) = −1. −1 = 0 + 5C2 ⇐⇒ C2 = −1

5.

Therefore, y(t) = cos(5t)− 1

5sin(5t).

S4.3 - Q36

y′′ − 4y′ + 13y = 0, y(0) = 4, y′(0) = 0. The characteristic equation is

λ2 − 4λ+ 13 = 0 ⇐⇒ λ =4±√

16− 4 · 13

2

⇐⇒ λ = 2± 3i.

A fundamental set of solutions is given by y1(t) = e2t cos(3t) and y2(t) = e2t sin(3t).

The general solution is y(t) = e2t(C1 cos(3t) + C2 sin(3t)

).

Put y(0) = 4. 4 = C1 + 0 ⇐⇒ C1 = 4.

Note y′(t) = e2t(

2C1 cos(3t) + 2C2 sin(3t)− 3C1 sin(3t) + 3C2 cos(3t))

.

=⇒ y′(t) = e2t(

8 + 3C2

)cos(3t) + e2t

(2C2 − 12

)sin(3t)

Put y′(0) = 0. 0 = 8 + 3C2 ⇐⇒ C2 = −8

3.

Therefore, y(t) = e2t(

4 cos(3t)− 8

3sin(3t)

).

S4.4 - Q11

We use the formula my′′ + ky = 0.

Here, m = 0.2 kg, k = 5N/m, y(0) = 0.5 m and y′(0) = 0 m/s.

0.2y′′ + 5y = 0 ⇐⇒ y′′ + 25y = 0, y(0) = 0.5, y′(0) = 0.

The characteristic equation is λ2 + 25 = 0 =⇒ λ = ±5i.

The general solution is y(t) = C1 cos(5t) + C2 sin(5t).

Put y(0) = 0.5. 0.5 = C1 + 0 ⇐⇒ C1 = 0.5.

Note y′(t) = −5C1 sin(5t) + 5C2 cos(5t).

Put y′(0) = 0. 0 = 0 + 5C2 ⇐⇒ C2 = 0.

Therefore, y(t) = 0.5 cos(5t).

The amplitude is A = 0.5 m. The phase is φ = 0. The frequency is ω0 = 5 rad/s.

S4.4 - Q14

mx′′ + kx = 0, x(0) = x0, x′(0) = v0.

Let ω20 =

k

m. We have x′′ + ω2

0x = 0.

The characteristic equation is λ2 + ω20 = 0 ⇐⇒ λ = ±iω0.

The general solution is x(t) = C1 cos(ω0t) + C2 sin(ω0t).

Put x(0) = x0. x0 = C1 + 0 ⇐⇒ C1 = x0.

Note x′(t) = −C1ω0 sin(ω0t) + C2ω0 cos(ω0t).

Put x′(0) = v0. v0 = 0 + C2ω0 ⇐⇒ C2 =v0ω0

.

Therefore, x(t) = x0 cos(ω0t) +v0ω0

sin(ω0t).

The amplitude is A =

√x20 +

( v0ω0

)2=

√x20 +

mv20k

.

S4.4 - Q16

In the first part, k =mg

x0=

1 · 9.84.9

= 2 N/m.

We use the formula my′′ + µy′ + ky = 0.

Here m = 1 kg, k = 2 N/m, µ = 3 kg/s, y(0) = −1 and y′(0) = −1.

y′′ + 3y′ + 2y = 0, y(0) = −1, y′(0) = −1.

The characteristic equation is λ2 + 3λ+ 2 = 0

⇐⇒ (λ+ 2)(λ+ 1) = 0 ⇐⇒ λ = −1 or − 2.

A fundamental set of solutions is given by y1(t) = e−t and y2(t) = e−2t.

The general solution is y(t) = C1e−t + C2e

−2t.

Put y(0) = −1. −1 = C1 + C2 · · · (Eq.1)

Note y′(t) = −C1e−t − 2C2e

−2t.

Put y′(0) = −1. −1 = −C1 − 2C2 · · · (Eq.2).

(Eq.1) + (Eq.2) : −2 = −C2 ⇐⇒ C2 = 2

=⇒ C1 = −1− C2 = −3.

Therefore, y(t) = −3e−t + 2e−2t.