first order differential equation (applications)

Applications of Linear Equations: Growth and Decay

Problems

The initial value problem

dx

dt = kx, x(t0) = x0 (1)

where k is a constant, occurs in many physical theories involving either

growth or decay. For instance, in biology it models the growth of bacteria or a small population of animals that increases at a rate proportional to the amount present at any time.

In physics the IVP (1) models the process of estimating the amount of

radioactive substance remaining at any time t or the temperature of a cooling

body. Similarly in chemistry, it may be used to estimate the amount os chemicals that remain at any time during certain chemical processes.

Bacterial Growth

Problem: A culture has N0 number of bacteria. At t = 1 hour the number of

bacteria is measured to be (3/2)N0. If the rate of growth is proportional to the

number of bacteria present, determine the time necessary for the number of bacteria to triple.

Answer: The associated initial value problem is

dN

dt = kN, N (0) = N0 (2)

which has the solution N (t) = N0ekt. Since N (1) = 3₂N0, we have 3

2N0 = N0e

k _{⇒ k = ln(}3

2) = 0.4055. Thus N (t) = N0e

0_.4055t.

Suppose that the bacteria population triples in time t = t1. Then

3N0 = N0e

0_.4055t1

⇒ t₁ = ln3

Carbon Dating

Problem: A fossilized bone is found to contain 1/1000 the original amount of

radioactive carbon C-14. Given the the half-life of C-14 in approximately 5600 years, determine the age of the fossil.

Answer: If A0 be the original amount of C-14, then the associated IVP is once

again of the same type as in the previous problem and it has the solution

A(t) = A0ekt. Since A(t) = A0/2 when t = 5600 years, we have

A0

2 = A0e

5600_t

⇒ k = − ln2

5600 = −0.00012378.

Therefore the process satisfies the equation, A(t) = A0e−0.00012378t.

Thus if the age of the fossil be t1 years, then

A0 1000 = A0e −0.00012378t1 _⇒ t 1 = ln1000 0.00012378 ≈ 55, 800 years.

Electrical Circuits

Problem: A 12-volt battery is connected to a simple series circuit in which the

inductance is 1/2 henry and the resistance is 10 ohms. Determine the current i

if the initial current is zero.

In a series circuit containing only a resistor and an inductor, Kirchoff’s second law states that the sum of the voltage drop across the inductor (Ldi_dt) and the voltage drop across the resistor (iR) is the same as the impressed voltage

(E(t)) on the circuit. Thus the current flow i(t) satisfies the linear equation

Ldi

dt + Ri = E(t) (3)

Thus the given problem gives rise to the IVP

1 2

di

dt + 10i = 12, i(0) = 0 (4)

Upon solving it we get the current flow as i(t) = 6₅ − 6

Fluid Mixtures

Problem: Initially 50 pounds of salt is dissolved in a large tank holding 300

gallons of water. A brine solution is pumped into the tank at a rate of 3 gallons

per minute and a well-stirred solution is then pumped out at the same rate. If the concentration of the solution entering is 2 pounds per gallon, determine

the amount of salt in the tank at any time. How much salt is present after 50

minutes? After a long time?

Answer: If A(t)be the amount of salt in the tank at any time t, then dA

dt = (Rate of entry) − (Rate of exit) = R1 − R2.

For the given problem the rate at which salt enters the tank is

R1 = (3gal/min) × (2lb/gal) = 6lb/min

and the rate at which salt leaves the tank is

R2 = (3gal/min) × (

A

300lb/gal) =

A

Fluid Mixtures

Thus the associated IVP is

dA

dt = 6 −

A

100, A(0) = 50 (5)

It has the solution A(t) = 600 − 550e−t/100. Thus at t = 50,

A(50) = 266.41 lbs. Also as t → ∞, we have A → 600. Thus after a long

period of time the amount of salt in the solution is 600 lbs.

If in the preceding example, the well stirred solution is pumped out at a slower rate of 2 gallons per minute, then the solution is accumulating at a rate of

(3 − 2)gal/min = 1 gal/min. After t minutes there are 300 + t gallons of brine

in the tank so that R2 = (2gal/min) ×



A

300+_tlb/gal



= ₃₀₀₊2A _tlb/min and

the IVP (5) takes the form

dA

dt +

2A

300 + t = 6, A(0) = 50, (6)

Applications of nonlinear equations:

Population

Growth

If the rate of growth in a population P is described by the equation dP

dt = kP, k > 0 then the population exhibits unbounded exponential

growth. This is an unrealistic model of growth in many cases, especially in those where the initial population is large. In such cases overcrowded

conditions with the resulting detrimental effects of such as pollution and excessive and competitive demand for resources inhibit growth. In 1840 a Belgian mathematician-biologist P. F. Verhulst proposed a different equation of the form

dP

dt = P (a − bP )

where a and b are positive constants determined by the circumstances. This

equation is referred to as the logistic equation and its solution as the logistic

Population Growth

In particular if a(> 0) is the average birthrate and the average death rate is

proportional to the population P (t) at any time t, then the rate of growth per

individual ( 1 P dP dt ) satisfies 1 P dP

dt = (average birth rate − average death rate) = a − bP.

Thus if P0 be the initial population, this gives rise to the IVP

dP

dt = aP − bP

2

, P (0) = P0 (7)

This has the solution

P (t) = aP0

bP0 + (a − bP0)e−at

.

Logistic curves are quite accurate predictors of growth patterns in a limited space of certain types of bacteria, protozoa, water fleas and fruit flies.

Spread of contagious diseases.

In the spread of contagious diseases, it is reasonable to expect that the rate

dx/dt at which the disease spreads is proportional not only to the number of

people x(t) that have already contracted disease, but also to those y(t) which

have not yet been affected. Thus

dx

dt = kxy

where k is the constant of proportionality.

Thus if one infected person is introduced into a population of n people, then

we have x + y = n + 1 so that the rate of spread of the disease is given by dx

dt = kx(n + 1 − x). This gives rise to the obvious initial value problem

dx

Spread of contagious diseases.

Problem: Suppose a student carrying a flu virus returns to an isolated college campus of 1000 students. Determine the number of infected students after 6

days if it is observed that 50 students are affected after 4 days.

Answer: Assuming that nobody leaves the campus throughout the duration of the disease, we seek the solution of the following IVP.

dx

dt = kx(1000 − x), x(0) = 1.

This has the solution x(t) = 1000

1 + 999e−1000kt. Using the fact that x(4) = 50

we have k = −1 4000ln 19 999 = 0.0009906 so that x(t) = 1000 1 + 999e−0.0009906t .

Thus the number of students affected after 6 days is

x(6) = 1000

Problems from mechanics: Falling bodies

It is well known that free-falling bodies close to the surface of the earth accelerate at a constant rate g. Since acceleration is the first derivative of

velocity which in turn is the first derivative of the distance s(t) covered in

time t, the vertical distance covered by the body is described by the equation

d2s

dt2 = g.

If v(t) be the velocity at any time t then d

2 s dt2 = dv dt = dv ds ds dt = v dv ds. Thus if

the body is falls from rest from a height h close to the earth’s surface, then this gives rise to the following first order IVP in v.

vdv

Falling bodies

If a falling body of mass m encounters air resistance which is proportional to

its velocity v(t) at any time time then the force acting on the body is mg − kv

where k is a constant of proportionality and the negative sign is due to the fact

that resistance opposes the motion.

Then Newton’s second law of motion implies that

mdv

dt = mg − kv.

Thus if the body falls from a height h, then this gives rise to the initial value

problem

vdv

ds + v

k

Projectiles

A rocket of mass m is propelled from the surface of the earth. its motion is

resisted by gravitational force which is inversely proportional to the square of the distance covered at any time t. Thus its motion is described by the

equation md 2_s dt2 = − k s2

where k is the constant of proportionality.

Using the fact that the acceleration due to gravity on the surface of the earth is

g if R be the radius of the earth, this yields the equation mg = k/R2_{. Thus if}

v0 be the initial velocity of the rocket, then the velocity v(s) at any height s is

the solution of the IVP

vdv

ds = −

gR2

Motion of a pendulum

A pendulum of mass m is fixed to the end of a rod of length a which has

negligible mass. If the bob is pulled to one side through an angle α and

released, then by the principle of conservation of energy.

1

2mv(t)

2

= mg(a cos α − a cos θ) where v(t)(= ds

dt ) is the

velocity at time t and θ is the angular displacement of the pendulum from the

vertical position at any time t. Since s = aθ, this gives rise to the IVP

1 2a dθ dt
2 = g(cos θ − cos α), θ(0) = α (12)

The shape of a hanging wire

Suppose that a suspended wire hangs under its own weight. Then the curve described by the shape of the wire is the solution of an IVP of second order which may be reduced to two IVPs of first order.

We examine a portion of the wire between the lowest point P1 and any other

point P2. This portion of the wire is at an equilibrium under the action of three

forces, namely, the weight of the segment P1P2, the tensions T1 and 2 in the

wire at P1 and P2 respectively. If w be the linear density of the wire and s the

The shape of a hanging wire

Resolving the tension T2 into its horizontal and vertical components T2 cos θ

and T2 sin θ respectively (see figure), the following equation arise from the

equilibrium condition.

T1 = T2 cos θ (13)

ws = T2 sin θ (14)

Since dx = ds cos θ, the length s of the arc P1P2 is given by

s = Z x 0 s 1 +  dy dx 2 (15)

From the fundamental theorem of integral calculus we have

ds dx = s 1 +  dy dx 2 (16)

The shape of a hanging wire

But from (13) and (14) we have tan θ = ws T1 so that dy dx = ws T1 (17)

Differentiating the above equation with respect to x we have d

2 y dx2 = w T1 ds dx.

Now using (16) we have

d2y dx2 = w T1 s 1 +  dy dx 2 .

Thus the shape of the wire is the solution of the IVP

d2y dx2 = w T1 s 1 +  dy dx 2 , y(0) = 0, dy dx = 0 when (x, y) = (0, 0) (18)

The shape of a hanging wire

Setting p = dy

dx, we get the IVP

dp

dx =

w T1

p1 + p2_{, p(0) = 0, (19).}

The solution of this IVP is the first order ODE dy

dx = tan

wx T1

. The latter gives rise to the second IVP

dy

dx = tan

wx T1

, y(0) = 0, (20).

Thus the shape of the wire obtained by solving this problem is

y = T1 w ln  sec wx T1  .