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.

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dy

y=

x

x 2 + 4( )dx

Second, integrate both sides.

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dy

y∫ =x

x 2 + 4( )dx∫

Solve for y.

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ln y =1

2ln x 2 + 4( ) +C1

Take e to both sides.

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y = eC1 x 2 + 4( )

or

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y = ±eC1 x 2 + 4( )

Ex. 2 Finding a Particular Solution

Given the initial condition y(0) = 1, find the particularsolution of the equation

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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.

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ex2

y

⎛

⎝ ⎜ ⎞

⎠ ⎟e−x 2

y 2 −1( )dy =ex

2

y

⎛

⎝ ⎜ ⎞

⎠ ⎟ −xydx( )

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y 2 −1( )

ydy = −xex

2

dx Now, integrate both sides.

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y 2 −1( )

ydy∫ = −xex

2

dx∫

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y −1

y

⎛ ⎝ ⎜

⎞ ⎠ ⎟dy∫ = −xex

2

dx∫

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y −1

y

⎛ ⎝ ⎜

⎞ ⎠ ⎟dy∫ = −xex

2

dx∫

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y 2

2− ln y =

u = x2

du = 2x dxdu/2x = dx

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−xeudu

2x∫

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−ex2

2+C Now find C at (0,1)

€

12

2− ln1 =

−e02

2+C

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1 = C

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y 2

2− ln y =

−ex2

2+1 or by multiplying by 2, you get

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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

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dy

dx=y

x 2

Now, separate the variables.

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dy

y∫ =dx

x 2∫

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ln y = −1

x+C1

Take e to both sidesto solve for y.

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y = e− 1/ x( )+C1

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y = eC1e− 1/ x( )

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y = Ce− 1/ x( )

at the point (1,3), C = ?

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3 = Ce− 1/1( )

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3e = C

So, the equation is

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y = 3e( )e−1/ x

or

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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.

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d N( )

dt= k 650 − N( ) Separate variables

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d(N) = k 650 − N( )dt Integrate

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dN

650 − N∫ = kdt∫ €

u'

u

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dN

650 − N= kdt

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−ln650 − N = kt +C1

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ln650 − N = −kt −C1

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ln650 − N = −kt −C1 Take e to both sides.

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650 − N = e−kt−C1

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N = 650 −Ce−kt

Using N = 300 when t = 0, you can conclude that C = 350,which produces

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N = 650 − 350e−kt

Then, using N = 500 when t = 2, it follows that

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500 = 650 − 350e−k(2)

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e−k(2) =3

7

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k ≈ 0.4236

So, the model for the coyote population is

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N = 650 − 350e−.4236t

When t = 3, the approximate population is

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N = 650 − 350e−.4236(3) ≈ 552 coyotes

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eC 1 = C