6.2 Solving Differential Equations

Modeling – Refers to finding a differential equation that describes a

given physical situation.

• Not too difficult, something I think is helpful is when you see y’ in an equation change it to dy/dx. (Leibniz notation)

• Process involves getting all x’s on one side, and all y’s on the other. We will treat dy/dx as a fraction and we can move the dy and dx with algebra.

Example

2' xyy

• In a lot of 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 shown.

dy kydt

Rate of change of y Proportional to yis

Constant of proportionality

'y ky dy kydt

dy k dty

dy k dty

1ln y kt C

1ln kt Cye e 1Ckty e e kty Ce

Leibniz Notation

Separation of variables

Integrate both sides

Natural log (on left side)

Solve for y

• 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.

• All solutions for y’=ky are of the form kty Ce

kty Ce

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

• When t = 0 and y = 2 (initial condition because time is 0)

kty Ce

02 2kty Ce Ce C 2

2

2

4 2

2

ln 2 lnln 2 2

ln 2 0.34662

k

k

k

e

e

ek

k

When t=2 and y=4

0.34662 tso y e0.3466(3)2 5.657so y e

• Suppose that 10 grams of the plutonium isotope Pu-239 was released in the Chernobyl nuclear accident. How long will it take 10 grams to decay to 1 gram? (remember this is decaying so k will be < 0)

• Let y be # of grams. Pu-239’s half life is 24,100 yearskty Ce

Initial conditions. t = 0, y = 10(0)

0

10

10 10

kCe

Ce C

(24,100)5 10 ke

Initial Conditions to find C.

Half life conditions

(24,100)5 10 ke24,1001

2ke

24,1001ln ln2

ke

ln(1/ 2) 24,100k12ln

24,100k

0.000028761k

0.00002876110 ty eFind time it takes to decay to 1 gram, then y=1

0.0000287611 10 te

80,059 years

Half Life Model

• Suppose an experimental population of fruit flies increases according to the law of exponential growth. There were 100 flies after the second day of the experiment and 300 flies after the fourth day. Approximately how many flies were in the original population?

• The first problem was not too difficult because the initial conditions were given, here the initial conditions are not given.

• Let y=Cekt be the number of flies at time t, where t is measured in days. Because y=200 when t=2 and y=300 when t=4, you can write 100=Ce2k and 300=Ce4k.

2100 kCe Take and solve for C. 22

100 100 kkC e

e

Now substitute the new C into the second equation.

2 4

2 4

2

300 100

300 100

3

k k

k k

k

e e

e

e

Substituting C into second equation and solving for k.

2ln(3) ln( )ln(3) 2

ln(3) 0.54932

kek

k

0.5493ty CeNow solve for C by reapplying the

condition that y=100 when t=2.

Then take that C and plug into the formula y=Cekt and plug in the initial time t=0 to find out the initial population of fruit flies.

• Newton’s Law of Cooling 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 (air most of the time).

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

' ( 60)y k y

Integrate both sides

( 60)dy k ydt

( 60)dy k y dt

60dy kdty

60dy k dty

1ln | 60 |y kt C 1ln| 60| kt Cye e

160 kt Cy e

60 kty Ce

Can omit the absolute value bars because y>60, its between 80 and 100.

60 kty Ce

0.0287760 40 ty e Use initial conditions of t=0 and y=100 to find C.

(0)100 60 40kCe C

Now when y=90 and t=10.

(10)90 60 40 ke 0.02877k

0.0287760 40 ty e (model)