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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