SECOND-ORDER DIFFERENTIAL EQUATIONS

SECOND-ORDER SECOND-ORDER DIFFERENTIAL EQUATIONSDIFFERENTIAL EQUATIONS

14

SECOND-ORDER DIFFERENTIAL EQUATIONS

Second-order linear differential

equations have a variety of applications

in science and engineering.

14.3Applications of Second-Order

Differential Equations

SECOND-ORDER DIFFERENTIAL EQUATIONS

In this section, we will learn how:

Second-order differential equations are applied

to the vibration of springs and electric circuits.

VIBRATING SPRINGS

We consider the motion

of an object with mass m

at the end of a spring that

is either vertical or

horizontal on a level

surface.

SPRING CONSTANT

In Section 6.4, we discussed Hooke’s Law:

If the spring is stretched (or compressed) x units from its natural length, it exerts a force that is proportional to x:

restoring force = –kx

where k is a positive constant, called the spring constant.

SPRING CONSTANT

If we ignore any external resisting forces

(due to air resistance or friction) then, by

Newton’s Second Law, we have:

This is a second-order linear differential equation.

2 2

2 2or 0

d x d xm kx m kxdt dt

Equation 1

SPRING CONSTANT

Its auxiliary equation is mr2 + k = 0 with

roots r = ±ωi, where

Thus, the general solution is:

x(t) = c1 cos ωt + c2 sin ωt

/k m

SIMPLE HARMONIC MOTION

This can also be written as

x(t) = A cos(ωt + δ)

where:

This is called simple harmonic motion.

2 21 2

1 2

/ frequency

amplitude

cos sin is the phase angle

k m

A c c

c c

A A

VIBRATING SPRINGS

A spring with a mass of 2 kg has natural

length 0.5 m.

A force of 25.6 N is required to maintain it

stretched to a length of 0.7 m.

If the spring is stretched to a length of 0.7 m and then released with initial velocity 0, find the position of the mass at any time t.

Example 1

VIBRATING SPRINGS

From Hooke’s Law, the force required to

stretch the spring is:

k(0.25) = 25.6

Hence, k = 25.6/0.2

= 128

Example 1

VIBRATING SPRINGS

Using that value of the spring constant k,

together with m = 2 in Equation 1,

we have:

2

22 128 0d x

xdt

Example 1

VIBRATING SPRINGS

As in the earlier general discussion,

the solution of the equation is:

x(t) = c1 cos 8t + c2 sin 8t

E. g. 1—Equation 2

VIBRATING SPRINGS

We are given the initial condition:

x(0) = 0.2

However, from Equation 2, x(0) = c1

Therefore, c1 = 0.2

Example 1

VIBRATING SPRINGS

Differentiating Equation 2, we get:

x’(t) = –8c1 sin 8t + 8c2 cos 8t

Since the initial velocity is given as x’(0) = 0, we have c2 = 0.

So, the solution is: x(t) = (1/5) cos 8t

Example 1

DAMPED VIBRATIONS

Now, we consider the motion of a spring that

is subject to either: A frictional force (the horizontal

spring here) A damping force (where a vertical

spring moves through a fluid, as here)

DAMPING FORCE

An example is

the damping force

supplied by a shock

absorber in a car or

a bicycle.

DAMPING FORCE

We assume that the damping force is

proportional to the velocity of the mass and

acts in the direction opposite to the motion.

This has been confirmed, at least approximately, by some physical experiments.

DAMPING CONSTANT

Thus,

where c is a positive constant,

called the damping constant.

damping force dxcdt

DAMPED VIBRATIONS

Thus, in this case, Newton’s Second Law

gives:

or 2

20

d x dxm c kxdt dt

2

2restoring force + damping force

d xmdt

dxkx c

dt

Equation 3

DAMPED VIBRATIONS

Equation 3 is a second-order linear

differential equation.

Its auxiliary equation is:

mr2 + cr + k = 0

Equation 4

DAMPED VIBRATIONS

The roots are:

According to Section 13.1, we need to discuss three cases.

2

1

2

2

4

2

4

2

c c mkr

m

c c mkr

m

Equation 4

CASE I—OVERDAMPING

c2 – 4mk > 0

r1 and r2 are distinct real roots.

x = c1er1t + c2er2t


Since c, m, and k are all positive,

we have:

So, the roots r1 and r2 given by Equations 4 must both be negative.

This shows that x → 0 as t → ∞.

2 4c mk c


Typical graphs of x as a function of f

are shown.

Notice that oscillations do not occur.

It’s possible for the mass to pass through the equilibrium position once, but only once.


This is because c2 > 4mk means that there

is a strong damping force (high-viscosity oil

or grease) compared

with a weak spring

or small mass.

CASE II—CRITICAL DAMPING

c2 – 4mk = 0

This case corresponds to equal roots

The solution is given by: x = (c1 + c2t)e–(c/2m)t

1 2 2

cr r

m

It is similar to Case I, and typical graphs

resemble those in the previous figure.

Still, the damping is just sufficient to suppress vibrations.

Any decrease in the viscosity of the fluid leads to the vibrations of the following case.

CASE II—CRITICAL DAMPING

CASE III—UNDERDAMPING

c2 – 4mk < 0

Here, the roots are complex:

where

The solution is given by:

x = e–(c/2m)t(c1 cos ωt + c2 sin ωt)

1

2 2

r ci

r m

24

2

mk c

m


We see that there are oscillations that are

damped by the factor e–(c/2m)t.

Since c > 0 and m > 0, we have –(c/2m) < 0. So, e–(c/2m)t → 0 as t → ∞.

This implies that x → 0 as t → ∞. That is, the motion decays to 0 as time increases.


A typical graph is shown.

DAMPED VIBRATIONS

Suppose that the spring of Example 1 is

immersed in a fluid with damping constant

c = 40.

Find the position of the mass at any time t if it starts from the equilibrium position and is given a push to start it with an initial velocity of 0.6 m/s.

Example 2

DAMPED VIBRATIONS

From Example 1, the mass is m = 2 and

the spring constant is k = 128.

So, the differential equation 3 becomes:

or

2

22 40 128 0d x dx

xdt dt

Example 2

2

220 64 0

d x dxx

dt dt

DAMPED VIBRATIONS

The auxiliary equation is:

r2 + 20r + 64 = (r + 4)(r + 16) = 0

with roots –4 and –12.

So, the motion is overdamped, and the solution is:

x(t) = c1e–4t + c2e–16t

Example 2

DAMPED VIBRATIONS

We are given that x(0) = 0.

So, c1 + c2 = 0.

Differentiating, we get:

x’(t) = –4c1e–4t – 16c2e

–16t

Thus, x’(0) = –4c1 – 16c2 = 0.6

Example 2

DAMPED VIBRATIONS

Since c2 = –c1, this gives:

12c1 = 0.6 or c1 = 0.05

Therefore, x = 0.05(e–4t – e–16t)

Example 2

FORCED VIBRATIONS

Suppose that, in addition to the restoring

force and the damping force, the motion

of the spring is affected by an external force

F(t).

FORCED VIBRATIONS

Then, Newton’s Second Law gives:

2

2restoring force damping force

external force

d xmdt

dxkx c F t

dt

FORCED VIBRATIONS

So, instead of the homogeneous equation 3,

the motion of the spring is now governed by

the following non-homogeneous differential

equation:

The motion of the spring can be determined by the methods of Section 13.2

Equation 5

2

2

d x dxm c kx F tdt dt

PERIOD FORCE FUNCTION

A commonly occurring type of external force

is a periodic force function

F(t) = F0 cos ω0t

where ω0 ≠ ω = /k m

PERIOD FORCE FUNCTION

In this case, and in the absence of a damping

force (c = 0), you are asked in Exercise 9 to

use the method of undetermined coefficients

to show that:

0

1 2 2 20

cos sin cos

x t

Fc t c t t

m

Equation 6

RESONANCE

If ω0 = ω, then the applied frequency

reinforces the natural frequency and

the result is vibrations of large amplitude.

This is the phenomenon of resonance.

See Exercise 10.

ELECTRIC CIRCUITS

In Sections 9.3 and 9.5, we were able to use

first-order separable and linear equations to

analyze electric circuits that contain a resistor

and inductor or a resistor and capacitor.

ELECTRIC CIRCUITS

Now that we know how to solve second-order

linear equations, we are in a position to

analyze this circuit.

ELECTRIC CIRCUITS

It contains in series:

An electromotive force E (supplied by a battery or generator)

A resistor R An inductor L A capacitor C

ELECTRIC CIRCUITS

If the charge on the capacitor at time t

is Q = Q(t), then the current is the rate of

change of Q with respect to t:

I = dQ/dt

ELECTRIC CIRCUITS

As in Section 9.5, it is known from physics

that the voltage drops across the resistor,

inductor, and capacitor, respectively,

are:dI Q

RI Ldt C

ELECTRIC CIRCUITS

Kirchhoff’s voltage law says that the sum of

these voltage drops is equal to the supplied

voltage:

dI QL RI E tdt C

ELECTRIC CIRCUITS

Since I = dQ/dt, the equation

becomes:

This is a second-order linear differential equation with constant coefficients.

2

2

1d Q dQL R Q E tdt dt C

Equation 7

ELECTRIC CIRCUITS

If the charge Q0 and the current I0 are known

at time 0, then we have the initial conditions:

Q(0) = Q0 Q’(0) = I(0) = I0

Then, the initial-value problem can be solved by the methods of Section 13.2

ELECTRIC CIRCUITS

A differential equation for the current can be

obtained by differentiating Equation 7 with

respect to t and remembering that I = dQ/dt:

2

2

1'

d I dIL R I E tdt dt C

ELECTRIC CIRCUITS

Find the charge and current at time t in

the circuit if:

R = 40 Ω L = 1 H C = 16 X 10–4 F E(t) = 100 cos 10t Initial charge and

current are both 0

Example 3

ELECTRIC CIRCUITS

With the given values of L, R, C, and E(t),

Equation 7 becomes:

E. g. 3—Equation 8

2

240 625 100cos10

d Q dQQ t

dt dt

ELECTRIC CIRCUITS

The auxiliary equation is r2 + 40r + 625 = 0

with roots

So, the solution of the complementary equation is:

Qc(t) = e–20t(c1 cos 15t + c2 sin 15t)

40 90020 15

2r i

Example 3

ELECTRIC CIRCUITS

For the method of undetermined coefficients,

we try the particular solution

Qp(t) = A cos 10t + B sin 10t

Then, Qp’ (t) = –10A sin 10t + 10B cos 10t

Qp’’(t) = –100A cos 10t – 100B sin 10t

Example 3

ELECTRIC CIRCUITS

Substituting into Equation 8, we have:

(–100A cos 10t – 100B sin 10t) + 40(–10A sin 10t + 10B cos 10t) + 625(A cos 10t + B sin 10t) = 100 cos 10t

or

(525A + 400B) cos 10t + (–400A + 525B) sin 10t = 100 cos 10t

Example 3

ELECTRIC CIRCUITS

Equating coefficients, we have:

525A + 400B = 100 –400A + 525B = 0

or

21A + 16B = 4 –16A + 21B = 0

The solution is: , 84697A 64

697B

Example 3

ELECTRIC CIRCUITS

So, a particular solution is:

The general solution is:

1697 84cos10 64sin10pQ t t t

201 2

4697

cos15 sin15

21cos10 16sin10

c p

t

Q t Q t Q t

e c t c t

t t

Example 3

ELECTRIC CIRCUITS

Imposing the initial condition Q(0),

we get:

841 697

841 697

0 0Q c

c

Example 3

ELECTRIC CIRCUITS

To impose the other initial condition, we first

differentiate to find the current:

201 2 1 2

40697

20 15 cos15 15 20 sin15

21sin10 16cos10

t

I

dQ

dt

e c c t c c t

t t

Example 3

ELECTRIC CIRCUITS

Thus,

6401 2 697

4642 2091

0 20 15 0I c c

c

Example 3

ELECTRIC CIRCUITS

So, the formula for the charge is:

20

63 cos 15 116 sin 1543

69721 cos 10 16 sin 10

t

Q t

et t

t t

Example 3

ELECTRIC CIRCUITS

The expression for the current is:

20

12091

1920 cos15 13,060 sin 15

120 21 sin 10 16 cos 10

t

I t

e t t

t t

Example 3

NOTE 1

In Example 3, the solution for Q(t) consists

of two parts.

Since e–20t → 0 as t → ∞ and both cos 15t and sin 15t are bounded functions,

204

2091 63 cos 15 116 sin 15 0

as

c

t

Q t

e t t

t

NOTE 1—STEADY STATE SOLUTION

So, for large values of t,

For this reason, Qp(t) is called the steady state solution.

4

697 21 cos 10 16 sin 10

pQ t Q t

t t

NOTE 1—STEADY STATE SOLUTION

The figure shows how the graph of the steady

state solution compares with the graph of Q /

in this case.

NOTE 2

Comparing Equations 5 and 7,

we see that, mathematically, they are

identical.

2

2

2

2

1

d x dxm c kx F tdt dt

d Q dQL R Q E tdt dt C

NOTE 2

This suggests the analogies given in the

following chart between physical situations

that, at first glance, are very different.

NOTE 2

We can also transfer other ideas from

one situation to the other.

For instance,

The steady state solution discussed in Note 1 makes sense in the spring system.

The phenomenon of resonance in the spring system can be usefully carried over to electric circuits as electrical resonance.

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SECOND-ORDER DIFFERENTIAL EQUATIONS