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6.3 Separation of Variables and the Logistic Equation. Ex. 1 Separation of Variables. Find the general solution of. First, separate the variables. y’s on one side, x’s on the other. Second, integrate both sides. Take e to both sides. Solve for y. or. - PowerPoint PPT Presentation
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6.3Separation of Variables and the Logistic Equation
Ex. 1 Separation of Variables
Find the general solution of
€
x 2 + 4( )dy
dx= xy
First, separate the variables.y’s on one side, x’s on the other.
€
dy
y=
x
x 2 + 4( )dx
Second, integrate both sides.
€
dy
y∫ =x
x 2 + 4( )dx∫
Solve for y.
€
ln y =1
2ln x 2 + 4( ) +C1
Take e to both sides.
€
y = eC1 x 2 + 4( )
or
€
y = ±eC1 x 2 + 4( )
Ex. 2 Finding a Particular Solution
Given the initial condition y(0) = 1, find the particularsolution of the equation
€
xydx + e−x 2
y 2 −1( )dy = 0
To separate the variables, you must rid the first termof y and the second term of e-x^2.
To do this, multiply both sides by ex^2/y.
€
ex2
y
⎛
⎝ ⎜ ⎞
⎠ ⎟e−x 2
y 2 −1( )dy =ex
2
y
⎛
⎝ ⎜ ⎞
⎠ ⎟ −xydx( )
€
y 2 −1( )
ydy = −xex
2
dx Now, integrate both sides.
€
y 2 −1( )
ydy∫ = −xex
2
dx∫
€
y −1
y
⎛ ⎝ ⎜
⎞ ⎠ ⎟dy∫ = −xex
2
dx∫
€
y −1
y
⎛ ⎝ ⎜
⎞ ⎠ ⎟dy∫ = −xex
2
dx∫
€
y 2
2− ln y =
u = x2
du = 2x dxdu/2x = dx
€
−xeudu
2x∫
€
−ex2
2+C Now find C at (0,1)
€
12
2− ln1 =
−e02
2+C
€
1 = C
€
y 2
2− ln y =
−ex2
2+1 or by multiplying by 2, you get
€
y 2 − ln y 2 + ex2
= 2
Ex. 3 Finding a Particular Solution Curve
Find the equation of the curve that passes through the point(1,3) and has a slope of y/x2 at any point (x,y).
Because the slope is y/x2, you have
€
dy
dx=y
x 2
Now, separate the variables.
€
dy
y∫ =dx
x 2∫
€
ln y = −1
x+C1
Take e to both sidesto solve for y.
€
y = e− 1/ x( )+C1
€
y = eC1e− 1/ x( )
€
y = Ce− 1/ x( )
at the point (1,3), C = ?
€
3 = Ce− 1/1( )
€
3e = C
So, the equation is
€
y = 3e( )e−1/ x
or
€
y = 3e −1/ x( )+1 = 3e x−1( ) / x
Ex. 4. Wildlife Population
The rate of change of the number of coyotes N(t) in a population is directly proportional to 650 - N(t), wheret is the time in years. When t = 0, the population is 300,and when t = 2, the population has increased to 500.Find the population when t = 3.
Because the rate of change of the population is proportionalto 650 - N(t), we can write the following differential equation.
€
d N( )
dt= k 650 − N( ) Separate variables
€
d(N) = k 650 − N( )dt Integrate
€
dN
650 − N∫ = kdt∫ €
u'
u
€
dN
650 − N= kdt
€
−ln650 − N = kt +C1
€
ln650 − N = −kt −C1
€
ln650 − N = −kt −C1 Take e to both sides.
€
650 − N = e−kt−C1
€
N = 650 −Ce−kt
Using N = 300 when t = 0, you can conclude that C = 350,which produces
€
N = 650 − 350e−kt
Then, using N = 500 when t = 2, it follows that
€
500 = 650 − 350e−k(2)
€
e−k(2) =3
7
€
k ≈ 0.4236
So, the model for the coyote population is
€
N = 650 − 350e−.4236t
When t = 3, the approximate population is
€
N = 650 − 350e−.4236(3) ≈ 552 coyotes
€
eC 1 = C