Chapter 5: Applications of Integration
Section 5.6: Differential Equations
Example 5.6.7
Solve the initial-value problem consisting of the logistic differential equation y./y=k c−y, and the initial condition y0=y0.
Solution
Mathematical Solution
The logistic differential equation models self-limiting population growth. Instead of a constant growth-rate per unit population, that constant rate is a linear function of the existing population. If the population falls below the fixed value c, the growth rate is positive and the population increases. If the population rises above the fixed value c, the growth rate is negative and the population decreases. If the population exactly equals the fixed value c, the growth rate is zero, and because y.=0, the population remains constant at the value y=c.
The value y=c is called the carrying capacity, and the solution y=c is called an equilibrium solution.
The logistic equation can be solved by separation of variables, but the resulting calculations - appearing in Table 5.6.7(a) - are not trivial. The algebra is much more tedious than the calculus! Step (2) in Table 5.6.7(a) is obtained by an algebraic technique called partial fractions, to be studied in detail in Chapter 6. It splits a fraction into "smaller" fractions that sum to the original; as such, it is the "opposite" of adding fractions over a common denominator. To verify that the left-hand side in (2) is equivalent to that in (1), simply add the fractions in (2).
In (3), absolute values are not needed in the logarithm of y because y>0. However, to allow y to be either greater than or less than c, the absolute value is retained. In step (8), the absolute value is dropped, but should so doing introduce a minus sign, that sign is incorporated into the arbitrary constant A^. To designate that A^ could now be either positive or negative, it is called A in all remaining steps in which it appears.
Separate variables
1.
dyy c−y
=k dt
2.
1cdyy+dyc−y
Integrate both sides
3.
lny−ln(c−y)
=c k t+λ
Solve for yt explicitly
4.
ln(yc−y)
5.
yc−y
=ec k t+λ
6.
=ec k t⋅eλ
7.
=A^ ec k t
8.
=A ec k t
9.
y
=c−yA ec k t
10.
y1+A ec k t
=c A ec k t
11.
=c A ec k t1+A ec k t
Apply the initial condition
12.
y0
=c A1+A
13.
1+A y0
=c A
14.
Ay0−c
=−y0
15.
A
=y0c−y0
16.
=cy0c−y0 ec k t1+y0c−y0 ec k t
17.
=c y0 ec k tc−y0+y0 ec k t
18.
=c y0c−y0e−c k t+y0
19.
=c y0y0+c−y0e−c k t
Table 5.6.7(a) Explicit solution of the IVP for the logistic equation
The typical text will display the solution of the IVP for the logistic equation in the form shown in (19).
Maple Solution
Control-drag the differential equation. Append the initial condition, making sure y0 is converted to the Atomic Identifier y__0.
Press the Enter key.
Context Panel: Solve DE≻yt
y./y=k c−y,y0=y__0
ⅆⅆtytyt=kc−yt,y0=y__0
→solve DE
yt=cy__0c−y__0ⅇ−kct+y__0
The "Collect a Function" dialog launches with the expression drop-down blank. However, the list is populated with one term, the exponential that appears in the expression where collection is to take place.
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