Section 6.2

Section 6.2

Differential Equations

A differential equation is an equation involving a derivative (slope).

A solution to a differential equation is a function whose derivative is the differential equation.

Example 1

Find the general solution to each differential equation.

a. yꞌ = 2 ' 2 dy

ydx2dy dx

2 dy dx

2 y x C

b. yꞌ + 2x2 = 0

22 0 dy

xdx

22dy

xdx

22dy x dx22 dy x dx

32

3 y x C

c. yyꞌ = -x

dy

y xdx

ydy xdx

ydy xdx

2 2

12 2

y xC

2 2 x y C

A family of solutions is infinite.

A particular solution is unique.

Example 2

Use integration techniques to find a general solution to

5 dy

x xdx

5 dy

x xdx

5 dy x x dx

5 dy x x dx

5 u x so du dx du dx

5 x u

1

25 y u u du

312 25 y u u du

3 52 2

53 5

2 2

u u

y C

312 25 y u u du

3 5

2 210 2

5 53 5

y x x C

Example 3

Find the particular solution to the differential equation yꞌ = cos x – sin x given y() = -1.

cos sin dy

x xdx

cos sin dy x x dx

sin cos y x x C

(, -1) 1 sin cos C

1 0 1 C0C

sin cos y x x Particular solution

General Solution

Example 4

Find the particular solution y = f(x) to the differential equation with the initial condition f(0) = −2.

Consider the differential equation 1

.

dy x

dx y

1

dy x

dx y

1 y dy x dx

2 21 1

2 2 y x x C General Solution

2 21 12 0 0

2 2 C

2 C

2 21 12

2 2 y x x

2 21 1

2 2 y x x C

2 2 2 4 y x x (0, −2)

2 2 4 y x x

Particular solution

Example 5

Solve the initial value problem:

2

22 6 , 0 1, ' 0 4

d yx y y

dx

2

22 6

d yx

dx

2 22 6d y x dx 22 3dy x x C dx

22 3dy

x x Cdx

24 2 0 3 0 C

4 C 22 3 4dy

x xdx

22 3 4dy

x xdx

22 3 4dy x x dx 2 3 4y x x x C

2 31 0 0 4 0 C

1 C3 2 4 1y x x x

6.2

2nd Day

Differential Equations: Growth and Decay

Growth and Decay Models

In many applications, the rate of change of a variable y is proportional to the value of y. If y is a function of time t, the proportion can be written as follows.

Solve the differential equation:

dyk dt

y

1 dy k dt

yln y kt C

y Cekt

So, all solutions of yꞌ = ky are in the form of y = Cekt

dy

kydt

kt Cy e e

ln y kt Ce e

Growth and Decay Models

If y is a differential function of t such that y > 0

and yꞌ = ky, for some constant k, then

y = Cekt

C is the initial value of y, and k is the

proportionality constant. Exponential growth

occurs when k > 0, and exponential decay

occurs when k < 0.

Example 1

The rate of change of y is proportional to y. When t = 0, y = 2, and when t = 2, y = 4. What is the value of y when t = 3?

Because y' = ky, you know that y and t are related by the equation y = Cekt.

You can find the value of the constant C by applying the initial conditions y = 2 when t = 0.

2 = Ce0

C = 2

You can find the value of the constant k by applying the initial conditions y = 4 when t = 2.

4 = 2e2k

2 = e2k

1ln 2

2k

0.3466k

ln 2 2 k

So, the model is y ≈ 2e0.3466t .

When t = 3,

y = 2e0.3466(3)

≈ 5.657

Notice that we did not actually have to solve the differential equation y' = ky, we just used its solution y = Cekt.

The next example concerns Newton's Law of Cooling, which states that:

the rate of change in the temperature of

an object is proportional to the difference between the object’s temperature and the temperature of the surrounding medium.

y′ = k(TO − TM)

( ) O M

dyk T T

dt

Example 2Let y represent the temperature (in ºF) of an object in a room whose temperature is kept at a constant 60º. If the object cools from 100º to 90º in 10 minutes, how much longer will it take for its temperature to decrease to 80º?

From Newton's Law of Cooling, you know that the rate of change in y is proportional to the difference between y and 60.

This can be written as

y' = k(y – 60), 80 ≤ y ≤ 100.

60 dy

k ydt

Solve: 60 dy

k ydt

1

60

dy k dt

y

1

60

dy k dty

ln 60 y kt C

ln 60 y kt C

ln 60 y kt C

60 kt Cy e

60 kty Ce

60 kty Ce

Using y = 100 when t = 0, then

Because y = 90 when t = 10,

90 = 60 + 40ek(10)

30 = 40e10k

0100 60 kCe40C60 40 kty e

1 3ln

10 4k

0.02877k

So, the Cooling Model is: y = 60 + 40e–0.02877t

and finally, when y = 80, you obtain

So, it will require about 14.09 more minutes for the object to cool to a temperature of 80º

0.0287780 60 40 te0.028771

2 te

1ln 0.02877

2 t

24.09 minutest

Half-Life Formula

Radioactive material decays exponentially and is measured in terms of half-life (the number of years required for half of the atoms in a sample of radioactive material to decay).

kty Ce

Example 3A sample contains 1 gram of radium. How much radium will remain after 1000 years? (Use a half-life of 1620 years.)

C = 1 grams at time t = 0.

kty Ce1 kty e

620,112

1 ke

ke1620ln2

1ln

k16202

1ln

.0004279k

1 kty e

.0004279(1000)1 y e

.652y g

.00042791 ty e

Example 4

Suppose that 10 grams of the plutonium isotope Pu-239 was released in the Chernobyl nuclear accident. How long will it take for the 10 grams to decay to 1 gram? Pu-239 has a half life of 24,100 years.

y = Cekt

10 kty e

1

2e24,100k

k 1

24,100ln

1

2

k .000028761

10 kty e 24,1005 10 ke

0.00002876110 ty e

0.0000287611 10 te

80,059 yearst

Example 5

Suppose a population of fruit flies increases according to the law of exponential growth (y = Cekt). There are 100 flies after Day 2 and 300 flies after Day 4. Approximately how many flies were in the original population?

kty Ce

(2,100) 2100 kCe

4300 kCe(4,300)

23 ke0.5493k

there were 33 flies to begin with.

2(.5493)100 Ce

Solve the system by dividing.

Example 6Money is deposited in an account for which

interest is compounded continuously. If the balance doubles in 6 years, what is the annual percentage rate?

A = Pert

A = Pert

2P = Pert

2 = e6r

ln 2 = 6r

.1155 11.55% r