first order differential equation (applications)

Applications of Linear Equations: Growth and Decay Problems The initial value problem dx = kx, x(t0 ) = x0 dt

(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.

Applications of First Order Ordinary Differential Equations – p. 1/1

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 = kN, N (0) = N0 dt

(2)

which has the solution N (t) = N0 ekt . Since N (1) = 32 N0 , we have 3 2 N0

= N0 ek ⇒ k = ln( 32 ) = 0.4055. Thus N (t) = N0 e0.4055t .

Suppose that the bacteria population triples in time t = t1 . Then 3N0 = N0 e0.4055t1 ⇒ t1 =

ln3 ≈ 2.71 hours . 0.4055

Applications of First Order Ordinary Differential Equations – p. 2/1

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) = A0 ekt . Since A(t) = A0 /2 when t = 5600 years, we have ln2 A0 = A0 e5600t ⇒ k = − = −0.00012378. 2 5600 Therefore the process satisfies the equation, A(t) = A0 e−0.00012378t . Thus if the age of the fossil be t1 years, then ln1000 A0 −0.00012378t1 = A0 e ≈ 55, 800 years. ⇒ t1 = 1000 0.00012378

Applications of First Order Ordinary Differential Equations – p. 3/1

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 di ) and the law states that the sum of the voltage drop across the inductor (L dt 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 L

di + Ri = E(t) dt

(3)

Thus the given problem gives rise to the IVP 1 di + 10i = 12, i(0) = 0 2 dt

(4)

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

6 5

− 65 e−20t .

Applications of First Order Ordinary Differential Equations – p. 4/1

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 = (Rate of entry) − (Rate of exit) = R1 − R2 . dt 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 A A lb/gal) = lb/min. R2 = (3gal/min) × ( 300 100

Applications of First Order Ordinary Differential Equations – p. 5/1

Fluid Mixtures Thus the associated IVP is dA A =6− , A(0) = 50 dt 100

(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 300 + t gallons of brine  there are 

A 2A lb/gal = 300+t lb/min and in the tank so that R2 = (2gal/min) × 300+t the IVP (5) takes the form 2A dA + = 6, A(0) = 50, (6) dt 300 + t This has the solution A(t) = 2(300 + t) − (4.95 × 107 )(300 + t)−2 .

Applications of First Order Ordinary Differential Equations – p. 6/1

Applications of nonlinear equations:

Population

Growth If the rate of growth in a population P is described by the equation dP = kP, k > 0 then the population exhibits unbounded exponential dt 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 = P (a − bP ) dt 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 function.

Applications of First Order Ordinary Differential Equations – p. 7/1

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 1 dP ) satisfies individual ( P dt 1 dP = (average birth rate − average death rate) = a − bP. P dt Thus if P0 be the initial population, this gives rise to the IVP dP = aP − bP 2 , P (0) = P0 dt This has the solution P (t) =

(7)

aP0 . −at bP0 + (a − bP0 )e

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

Applications of First Order Ordinary Differential Equations – p. 8/1

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 = kxy dt 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 = kx(n + 1 − x). This gives rise to the obvious initial value problem dt dx = kx(n + 1 − x), x(0) = 1. dt

(8)

Applications of First Order Ordinary Differential Equations – p. 9/1

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 = kx(1000 − x), x(0) = 1. dt 1000 . Using the fact that x(4) = 50 −1000kt 1 + 999e 19 1000 −1 ln = 0.0009906 so that x(t) = we have k = . 4000 999 1 + 999e−0.0009906t Thus the number of students affected after 6 days is This has the solution x(t) =

1000 = 276 students. x(6) = 1 + 999e−5.9436

Applications of First Order Ordinary Differential Equations – p. 10/1

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 d2 s = g. dt2

dv ds dv dv d2 s = = v . Thus if If v(t) be the velocity at any time t then 2 = dt dt ds dt ds 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. dv = g, v(h) = 0 v ds

(9)

Applications of First Order Ordinary Differential Equations – p. 11/1

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 m

dv = mg − kv. dt

Thus if the body falls from a height h, then this gives rise to the initial value problem k dv + v = g, v(h) = 0 (10) v ds m

Applications of First Order Ordinary Differential Equations – p. 12/1

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 k d2 s m 2 =− 2 dt s 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/R 2 . 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 gR2 dv = − 2 , v(R) = v0 v ds s

(11)

Applications of First Order Ordinary Differential Equations – p. 13/1

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.

ds ) is the = mg(a cos α − a cos θ) where v(t)(= dt 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 dθ 2 = g(cos θ − cos α), θ(0) = α (12) 2 a dt 1 2 2 mv(t)

Applications of First Order Ordinary Differential Equations – p. 14/1

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 P1 P2 , 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 length of P1 P2 , then the weight of P1 P2 is ws.

Applications of First Order Ordinary Differential Equations – p. 15/1

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 P1 P2 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)

Applications of First Order Ordinary Differential Equations – p. 16/1

The shape of a hanging wire But from (13) and (14) we have tan θ =

ws so that T1

ws dy = dx T1

(17)

w ds d2 y . Differentiating the above equation with respect to x we have 2 = dx T1 dx Now using (16) we have w d2 y = dx2 T1

s

1+



dy dx

2

.

Thus the shape of the wire is the solution of the IVP 2

w d y = dx2 T1

s

1+



dy dx

2

dy = 0 when (x, y) = (0, 0) , y(0) = 0, dx

(18)

Applications of First Order Ordinary Differential Equations – p. 17/1

The shape of a hanging wire dy , we get the IVP Setting p = dx wp dp = 1 + p2 , p(0) = 0, dx T1

(19).

The solution of this IVP is the first order ODE

wx dy = tan . The latter gives dx T1

rise to the second IVP wx dy = tan , y(0) = 0, dx T1

(20).

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



wx T1 ln sec w T1



.

Applications of First Order Ordinary Differential Equations – p. 18/1