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Applications of First Order Differential Equations

First order differential equations have an applications in Electrical circuits, growth and decay problems, temperature and falling body problems and in many other fields. I’ll discuss here some of them one by one.

Let N(t) denote the amount of substance (or population) that is either growing or decaying. If we assume that dN/dt, the time rate of change of this amount of substance, is proportional to the amount of substance present. Then

dN/dt = kN

or

dN/dt - KN = 0 (1)
where k is the constant of proportionality, and we assume that N(t) is a differential.

A person places $20,000 in a savings account which pays 5 percent interest per annum, compounded continuously. Find (a) the amount in the account after three years, and (b) the time required for the account to double in value, presuming no withdrawals and no additional deposits.

2. This differential equation is both linear and separable. Its solution is

(2)

3. At t= 0, N(0) = 20,000, which when substituted into (2) yields

4. With this value of c, (2) becomes

(3)

Equation (3) gives the dollar balance in the account at any time t.

5. (a) Substituting t = 3 into (3), we find the balance after three years to be

6. (b) We seek the time t at which N(t) = $40,000. Substituting these values into (2) and solving for t, we obtain

Newton’s Law of cooling, which is equally applicable to heating, states that the time rate of change of the

temperature of a body is proportional to the temperature difference between the body and its surrounding medium. Let T denote the temperature of the body and let Tm denote the temperature of the surrounding medium. Then the time rate of change of Ihe temperature of the body dT/dt, and Newton’s law of cooling can

be formulated as

dT/dt = -k(T- Tm)

or as

dT/dt +kT=kTm (4)
where k is a positive constant of proportionality. Once k is chosen positive, the minus sign is required in Newton’s law to make dT/dt negative in a cooling process, when T is greater than Tm. and positive in a heating process, when T is less than Tm.

A metal bar at a temperature of 100° F is placed in a room at a constant temperature of 0°F. If after 20 minutes the temperature of the bar is 50° F, find (a) the time it will take the bar to reach a temperature

of 25° F and (b) the temperature of the bar after 10 minutes.

2. This equation is linear and its solution is

(5)

3. Since T= 100 at t = 0 (the temperature of the bar is initially 100° F), it follows from (5) that 100 = ce-k(0) or 100 =c. Substituting this value into (5), we obtain

(6)

4. At t= 20, we are given that T= 50, hence, from (6)

5. Substituting this value into (6), we obtain the temperature of the bar at any time t as

(7)

6. (a) We require t when T = 25. Substituting T=25 into (7), we have

Solving, we find that t = 39.6 min.

7. (b) We require T when t = 10. Substituting t= 10 into (7) and then solving for T, we find that

We can analyze the series RC and RL circuits using first order differential equations.

For series RL circuit as shown in figure (1) consisting of resistance R, inductance L and voltage source E, differential equation is of the form

(8)

Fig. (1)

For series RC circuit as shown in figure (2) consisting of resistance R, capacitance C and voltage source E, differential equation is of the form

(9)

where q represents the charge stored in a capacitor, and current I=dq/dt.

Fig. (2)

An RL circuit has an emf of 5 volts, a resistance of 50 ohms, an inductance of 1 henry, and no initial

current. Find the current in the circuit at any time t.

2. This is first order linear differential equation, its solution is

3. Substituting t=0, I=0 ,we get c=1/10, thus current at any time t is

The term (-e-50t/10) is the transient current and 1/10 is the steady state current.

An RC circuit has an emf given (in volts) by 400 cos 2t, a resistance of 100 ohms, and a capacitance of

10^-2 farad. Initially there is no charge on the capacitor. Find the current in the circuit at any time t.

1. Find the charge q, Here, E = 400cos2t, R= 100, and

C= 10^-2 , hence (9) becomes

2. This equation is linear and its solution is

3. Substituting t=0,q=0 in above equation to get

4. Then find current as

Consider a vertically falling body of mass m that is being influenced only by gravity g and an air resistance that is proportional to the velocity of the body. Assume that both gravity and mass remain constant and, for convenience, choose the downward direction as the positive direction.

Newton’s second law of motion:

“ The net force acting on a body is equal to the time rate of change of the momentum of the body, or, for constant mass”.

(10)

where F is the net force on the body and v is the velocity of the body, both at time t.

For the problem in figure (3), there are two forces acting on the body: (1) the force due to gravity given by the weight w of the body, which equals mg, and (2) the force due to air resistance given by -kv, where k > 0 is a constant of proportionality. The minus sign is required because this force opposes the velocity; that is, it acts in

the upward, or negative, direction. The net force F on the body is, therefore

F = mg-kv.

Substituting this result into (10), we obtain

(11)

as the equation of motion for the body.

If air resistance is negligible, then k = 0 and equation (11) simplifies to

Fig. (3)

A body of mass 5 slugs is dropped from a height of 100 ft with zero velocity. Assuming no air resistance, find (a) an expression for the velocity of the body at any time t, (b) an expression for the position of the body at any time t, and (c) the time required to reach the ground.

(a) since there is no air resistance, dv/dt = g applies. This

differential equation is linear or, in differential form, separable; its solution is v = gt+c. When t=0, v = 0 (initially the body has zero velocity); hence 0 = g(0) + c, or c = 0. Thus, v = gt or, assuming g = 32 ft/sec2
.

v=32t (12)

(b) Recall that velocity is the time rate of change of displacement x. Hence, v = dx/dt, and

(12) becomes dx/dt = 32t. This differential equation is also both linear and separable; its solution is

x=16t2 +c1 (13)

Substituting t = 0, x=0 to get c1=0, and thus equation (13) becomes

x=16t2 (14)

(c) We require t when x = 100. From (14) t = ((100)/(16))1/2= 2.5sec.

A person places $5000 in an account that accrues interest compounded continuously. Assuming no additional deposits or withdrawals, how much will be in the account after seven years if the interest rate is a constant 8.5 percent for the first four years and a constant 9.25 percent for the last three years?

An RL circuit has an emf given (in volts) by 3 sin 2t, a resistance of 10 ohms, an inductance of 0.5 Henry, and an initial current of 6 amperes. Find the current in the circuit at any time t.

A steel ball weighing 2 Ib is dropped from a height of 3000 ft with no velocity. As it falls, the ball encounters air resistance numerically equal to v/8 (in pounds), where v denotes the velocity of the ball (in feet per second). Find (a) the limiting velocity for the ball and (b) the time required for the ball to hit the ground

A body at a temperature of 50° F is placed outdoors where the temperature is 100° F. If after 5 minutes
,the temperature of the body is 60° F, find (a) how long it will take the body to reach a temperature of 75° F and (b) the temperature of the body after 20 minutes.

Reference:

Schaum’s outline of differential equations by Gabriel B. Costa and Richard Bronson.

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