MATH 42 — Solving the Logistic Equation
Here is how to solve the logistic equation
y
(
t
) =
k
1

y
(
t
)
K
y
(
t
)
using the method of separable equations.
First, use the general ideas (move everything except the
k
over to the left hand side, integrate
both sides from 0 to
T
, change variables on the left hand side) as well as partial fractions to get
kT
=
y
(
T
)
y
(0)
du
(1

u/K
)
u
=
y
(
T
)
y
(0)
1
u
+
1
K

u
du
= ln
u
K

u
y
(
T
)
y
(0)
= ln
y
(
T
)(
K

y
(0))
(
K

y
(
T
))
y
(0)
.
Next set
A
= (
K

y
(0))
/y
(0) for convenience and take the exponential of both sides, which gives
e
kT
=
Ay
(
T
)
K

y
(
T
)
.
Next we’ll isolate
y
(
T
) by clearing denominators and solving the resulting linear equation for
y
(
T
).
Note there’s a bit of sloppiness taking place here. Read this paragraph only if you’re losing sleep
because we’ve lost the absolute value sign! This happened for the following reasons... Technically
speaking we need to assume
y
(0) =
K
and
y
(0) = 0 for the expression above to make sense. (And
it is clear that there are solutions of the logisitic equation satisfying
y
(0) =
K
and
y
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 Winter '07
 Staff
 Linear Algebra, Algebra, Equations, Differential Calculus, Elementary algebra, Continuous function, one hand, logistic equation

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