section 6.2
DESCRIPTION
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 . y ꞌ = 2. - PowerPoint PPT PresentationTRANSCRIPT
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