6.1/6.2 Differential Equations – separating

variablesBy Tessa Davidson

A differential equation will be in the form of

In order to solve it, you must put it in the form of

To solve, you take the integral of both sides to get

( ) ( )dy

f x g ydx

( ) ( )g y dy f x dx

( )y h x

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.

Exponential Growth and Decay Model – 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.

y = Cekt

( ) ktdTk T S T Ce S

dt

Example: 5dy x

dx y

Example: 5dy x

dx y

5ydy xdx

Example: 5dy x

dx y

5ydy xdx

5ydy xdx

Example: 5dy x

dx y

5ydy xdx

5ydy xdx 2 2

1 2

5

2 2

y xc c

Example: 5dy x

dx y

5ydy xdx

5ydy xdx 2 2

1 2

5

2 2

y xc c

2 25y x C

Example: 4u tdue

dt

Example: 4u tdue

dt 4u tdu

e edt

Example: 4u tdue

dt 4u tdu

e edt

41 tudu e dt

e

Example: 4u tdue

dt 4u tdu

e edt

41 tudu e dt

e 4u te du e dt

Example: 4u tdue

dt 4u tdu

e edt

41 tudu e dt

e 4u te du e dt

41 2

1

4u te c e c

Example: 4u tdue

dt 4u tdu

e edt

41 tudu e dt

e 4u te du e dt

41 2

1

4u te c e c

41

4u te e C

General Solution -

Particular Solution -

In order to find a particular solution, you must first have an initial condition.

xy Ce

6 xy e

We get this general solution from the equation

1

1

'

'

'

1

lnCkt

kt

y ky

yk

y

ydt kdt

y

dy kdty

y kt C

y e e

y Ce

Finding a particular solutionGiven the initial condition , find the

particular solution of the equation

Note that is a solution of the differential equation—but this solution does not satisfy the initial condition. So, you can assume that . To separate variables, you must rid the first term of y and the second term of . So, you should multiply by and obtain the following.

(0) 1y 2 2( 1) 0xxydx e y dy

0y

0y

2xe 2

/xe y

2 2( 1) 0xxydx e y dy 2

2

2

2

2

( 1)

1( )

1ln 2

2 2

x

x

x

e y dy xydx

y dy xe dxy

ye C

Exponential Growth and Decay

Radioactive DecaySuppose 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 1 gram?

Solution: let y represent the mass ( in grams) of the plutonium. Because the rate of decay is proportional to y, you know that

where t is the time in years. To find the values of the constants C and k, apply the initial conditions. Using the fact that y=10 when t=0, you can write

kty Ce

(0) 010 kCe Ce

Which implies that c=10. Next, using the fact that y=5 when t=24,100 you can write

So, the model is

(24,100)

24,100

5 10

1

21 1

ln24,100 2

0.000028761

k

k

e

e

k

k

0.00002876110 ty e

To find the time it would take for 10 grams to decay to 1 gram, you can solve for t in the equation

The solution is approximately 80,059 years

0.0000287611 10 te