Deriving general solution for logistic differential equation

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warwick

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I'm starting to review for the differential equations and vector analysis courses I'm taking in the fall. It's been awhile since I finished Cal 3. I took Linear Algebra this past fall. Any other tips and advice would be appreciated. Down to the question:

In the book, they derive

dy/dt = ky (1 - y/L)

They rewrite as partial fractions, and I'm not seeing how.

1 / (y (1 - y/L))

as

1/y + 1/(L-y)

I get those factors, but I'm not seeing how they solved for "A" and "B" in the partial fractions technique.
 
L is a constant, so treat it as such.

\(\displaystyle \frac{1}{y(1-\frac{y}{L})}=\frac{L}{y(L-y)}\)

Try it this way instead of all those A's and B's:

Add y and subtract y:

\(\displaystyle \frac{L-y+y}{y(L-y)}\)

\(\displaystyle \frac{L-y}{y(L-y)}+\frac{y}{y(L-y)}\)

\(\displaystyle \frac{1}{y}+\frac{1}{L-y}\)
 
Another way:

L/[y(L - y)] = A/y + B/(L - y)

Multiply both sides by the LCD (y(L - y)):

L = A(L - y) + By

If y = L, L = BL, so B = 1.

If y = 0, L = -AL, so A = -1.

L/[y(L - y)] = -1/y + 1/(L - y)
 
Galactus, will that method work in many cases in place of the "A"s and "B"s.

Fast eddie, when I first worked it, I got 1=BL and 1=AL. That didn't really tell me much, but having that L in the numerator made the difference.

Thanks for your help. All these little tricks and techniques are slowly coming back to me. lol.
 
Galactus, will that method work in many cases in place of the "A"s and "B"s.
Click to expand...

Yes. If you are good enough at algebra you can do it anytime in place of the A, B, C thing. Depending on how complicated the expression is.


Here's one slightly more complicated, if you like. Try expanding \(\displaystyle \frac{3}{n^{3}-1}\)

Of course, we can factor the denominator using the difference of two cubes thing:

\(\displaystyle \frac{3}{(n-1)(n^{2}+n+1)}\)

Now, manipulating the top by adding and subtracting, but note it still equals 3.

\(\displaystyle \frac{-n^{2}+n^{2}-2n+2n+2+1}{(n-1)(n^{2}+n+1)}\)

Maneuver things around a bit:

\(\displaystyle \frac{-2n+2+n^{2}+n+1-n^{2}-n}{(n-1)(n^{2}+n+1)}\)

\(\displaystyle \frac{-2(n-1)+(n^{2}+n+1)-n(n-1)}{(n-1)(n^{2}+n+1)}\)

\(\displaystyle \frac{-2}{n^{2}+n+1}+\frac{1}{n-1}-\frac{n}{n^{2}+n+1}\)

There she is.
 
\(\displaystyle The \ Analytic \ Solution.\)

\(\displaystyle \frac{dy}{dt} = ky(1-\frac{y}{L}) = ky(\frac{L-y}{L})\)

\(\displaystyle \int \frac{L \ dy}{y(L-y)} = \int kdt\)

\(\displaystyle \int (\frac{1}{y}+\frac{1}{L-y})dy = kt +C\)

\(\displaystyle ln(y)-ln(L-y) = kt +C\)

\(\displaystyle ln(L-y)-ln(y) = -kt-C\)

\(\displaystyle ln|\frac{L-y}{y}| = -kt-C\)

\(\displaystyle |\frac{L-y}{y}| = (e^{-kt})(e^-C})\)

\(\displaystyle \frac{L-y}{y} = Ae^{-kt}, \ A =\pm e^{-C}.\)

\(\displaystyle \frac{L}{y}-1 = Ae^{-kt}, \ \frac{L}{y} = Ae^{-kt}+1\)

\(\displaystyle \frac{y}{L} = \frac{1}{Ae^{-kt}+1}\)

\(\displaystyle y = \frac{L}{Ae^{-kt}+1}\)

\(\displaystyle Now \ if \ t=0, \ then \ y = y_\sub0 \ (the \ initial \ amount), \ so \ \frac{L-y_\sub0}{y_\sub0} = Ae^{0} = A.\)

\(\displaystyle Then \ the \ solution \ to \ the \ logistic \ equation \ is:\)

\(\displaystyle y(t) = \frac{L}{1+Ae^{-kt}} \ where A = \frac{L-y_\sub0}{y_\sub0}.\)
 
galactus said:
Galactus, will that method work in many cases in place of the "A"s and "B"s.
Click to expand...

Yes. If you are good enough at algebra you can do it anytime in place of the A, B, C thing. Depending on how complicated the expression is.


Here's one slightly more complicated, if you like. Try expanding \(\displaystyle \frac{3}{n^{3}-1}\)

Of course, we can factor the denominator using the difference of two cubes thing:

\(\displaystyle \frac{3}{(n-1)(n^{2}+n+1)}\)

Now, manipulating the top by adding and subtracting, but note it still equals 3.

\(\displaystyle \frac{-n^{2}+n^{2}-2n+2n+2+1}{(n-1)(n^{2}+n+1)}\)

Maneuver things around a bit:

\(\displaystyle \frac{-2n+2+n^{2}+n+1-n^{2}-n}{(n-1)(n^{2}+n+1)}\)

\(\displaystyle \frac{-2(n-1)+(n^{2}+n+1)-n(n-1)}{(n-1)(n^{2}+n+1)}\)

\(\displaystyle \frac{-2}{n^{2}+n+1}+\frac{1}{n-1}-\frac{n}{n^{2}+n+1}\)

There she is.
Click to expand...

Galactus,

How did you know what to add and subtract in the denominator? Is it a trial-and-error process?

Also, wouldn't you have to do the same thing or another technique for the first and third terms in your partial fractions simplification?

BigGlenn,

Thanks. I really only had trouble with the Partial Fractions technique. I'm going over Parts now.,
 
Yes, what I done there was more like just having done so many over the years and recognizing the form

Note, I added and subtracted the largest terms from the denominator.

I would just suggest doing it the standard way.
 
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