APPLICATIONS OF DIFFERENTIAL EQUATIONS

In this section we solve problems on differential equations which have direct impact on real life situation. Solving of these types of problems involve

(i) Construction of the mathematical model describing the given situation.

(ii) Seeking solution for the model formulated in (i) using the methods discussed earlier.

Illustration :

Let A be the amount a substance at any time t. The rate of change of the substance is directly proportional to the the amount of substance exists at that time.

That is,

dA/dt ∝ A

dA/dt = kA

where k is called the constant of proportionality.

1. If k > 0, we say that A grows exponentially with growth constant k (growth problem).

2. If k < 0 we say that A decreases exponentially with decreasing constant k (decay problem).

In all the word problems, we apply the principle that the rate of change of substance is directly proportional to the amount of substance exists at that time.

That is,

dA/dt ∝ A

dA/dt = kA

Here k may be positive or negative depends on the problem.

Solving the differential equation dA/dt = kA for A :

dA/dt = kA

Multiply both sides by dt.

dA = kAdt

Divide both sides by dt.

dA/A = kdt

Take the integral on both sides.

∫dA/A = ∫kdt

lnA = kt + logC

Convert the above equation to exponential.

A = e^{kt + logC}

A = e^logCe^kt

A = Ce^kt

If t = 0,

A = Ce^k(0)

A = Ce⁰

A = C(1)

A = C

If t = 0 (initial time), then the amount substance A = C.

So, C stands for the initial amount of substance, that is, when t = 0.

When you solve a word problem using differential equation, your first work is to solve for k in A = Ce^kt using the information given in the question.

Example 1 :

Radioactive radium has a half-life of approximately 1599 years.

A. If 50 milligrams of pure radium is present initially, when will the amount remaining be 20 milligrams? Give your answer to the nearest year.

B. What percent of a given amount remains after 100 years?

Solution :

Part A :

Let A be the amount of radium exists after t years of time.

A = Ce^kt

Given : 50 milligrams of pure radium is present initially.

Then, C = 50.

A = 50e^kt----(1)

Solve for k :

It is given that half-life of approximately 1599 years.

Then, A = 25, when t = 1599.

25 = 50e^k(1599)

0.5 = e^1599t

Convert the above equation to logarithmic form.

ln0.5 = 1599k

ln0.5/1599 = k

Target : To find the time when A = 20.

In (1) substitute A = 20.

20 = 50e^kt

0.4 = e^kt

Convert the above equation to logarithmic form.

ln0.4 = kt

Substitute k = ln0.5/1599.

ln0.4 = (ln0.5/1599)t

(ln0.4 x 1599)/ln0.5 = t

Substitute ln0.4 = -0.9163 and ln0.5 = -0.6931.

(-0.9163 x 1599)/(-0.6931) = t

t ≈ 2114 years

20 milligrams of radium will be remaining in about 2114 years.

Part B :

To know the percent of a given amount remains after 100 years, substitute t = 100 in (1).

A = 50e^k(100)

A = 50e^100k ----(2)

Find the value of 100k :

100k = 100 x ln0.5/1599

= 100 x (-0.6931)/1599

= -69.31/1599

= -0.0433

Substitute 100k = -0.0433A in (2).

A = 50e^-0.0433

= 50(1/e^0.0433)

Substitute e ≈ 2.71828.

= 50(1/2.71828^0.0433)

= 50(1/1.0443)

= 50 x 0.957

= 0.957 x 50

= (95.7/100) x 50

= 95.7% of 50.

95.7 percent of the given amount remains after 100 years.

Example 2 :

Water is leaking out of a large barrel such that the rate of the change in the water level is proportional to the square root of the depth of the water at that time. If the water level starts at 36 inches and drops to 34 inches in 1 hour, how long will it take for all of the water to drain out of the barrel?

Solution :

Part A :

Let D be the depth of the water at any time t.

When t = 0, D = 36.

When t = 1, D = 34.

If all of the water drains out of the barrel, then D = 0.

So, we have to find the value of t when D = 0.

It will take about 35.5 hours for for all of the water to drain out of the barrel.

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APPLICATIONS OF DIFFERENTIAL EQUATIONS

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