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AKG
AKG is offline
#2
Jan25-05, 11:33 PM
Sci Advisor
HW Helper
P: 2,589
Separation of Variables:

[tex]\frac{dY}{dx} = \left (\frac{a}{b}\right ) \left (Y(1 - Y)\right )[/tex]

[tex]\frac{dY}{Y(1 - Y)} = \frac{a}{b}dx[/tex]

[tex]\int \frac{dY}{Y(1 - Y)} = \int \frac{a}{b}dx[/tex]

[tex]\int \left (\frac{A}{Y} + \frac{B}{1 - Y}\right )dY = \frac{a}{b}x + C[/tex]

[tex]\int \left (\frac{A}{Y} + \frac{B}{1 - Y}\right )dY = \frac{a}{b}x + C[/tex]

where A and B are constants that need to be solved for. It's easy to see that A = B = 1, so:

[tex]\int \left (\frac{1}{Y} - \frac{1}{Y - 1}\right )dY = \frac{a}{b}x + C[/tex]

[tex]\ln (Y) - \ln (Y - 1) = \frac{a}{b}x + C[/tex]

[tex]\ln \right (\frac{Y}{Y - 1}\right ) = \frac{a}{b}x + C[/tex]

[tex]\frac{Y}{Y - 1} = \exp \left (\frac{a}{b}x + C\right )[/tex]

[tex]\frac{Y}{Y - 1} = De^{\frac{a}{b}x}[/tex]

where [itex]D = e^C[/itex]. Some algebra gets you to:

[tex]Y = \frac{D}{D - \exp (-\frac{ax}{b})}[/tex]

There may be some cases where this solution is not valid, i.e. in the steps above, I may have divided by zero if Y = 1 or Y = 0 in some places, you can check the algebra. In fact, the above solution only holds in the case that Y is neither 0 or 1, since Y = 0 and Y = 1 are solutions on their own. Note:

If Y = 0 or Y = 1, it is a constant, so dY/dx = 0. Also, it is clear that aY(1-Y)/b = 0 for those Y values, as we would expect.

Now, you say that you have, "when X = 0, Y = something." This is an initial value problem. With this, you can solve for D explicitly (in terms of a, b, and other constants).