Logistic growth model, differential equation

  • Thread starter Wiz14
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  • #1
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Homework Statement


dY/dt = y(c - yb)

C and B are constants.
Im supposed to find and explicit solution for y, but im having trouble.

Homework Equations





The Attempt at a Solution



dY/y(c - yb) = dt
∫(1/c)dy/y + ∫(b/c)dY/c - yb = ∫dt (i used partial fraction decompositions)
(1/c)ln|y| - (b/c)ln|c - yb| = t + K (K stands for an arbitrary constant)
ln|[(c-yb)^b]/y| = -ct + K (Multiplied by negative c and then combine the natural logs on the left)

What im having trouble with is the last expression on the LHS. I have (c-yb)^b so how am I supposed to solve for y, if i dont know b?
 

Answers and Replies

  • #2
Curious3141
Homework Helper
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87

Homework Statement


dY/dt = y(c - yb)

C and B are constants.
Im supposed to find and explicit solution for y, but im having trouble.

Homework Equations





The Attempt at a Solution



dY/y(c - yb) = dt
∫(1/c)dy/y + ∫(b/c)dY/c - yb = ∫dt (i used partial fraction decompositions)
(1/c)ln|y| - (b/c)ln|c - yb| = t + K (K stands for an arbitrary constant)
ln|[(c-yb)^b]/y| = -ct + K (Multiplied by negative c and then combine the natural logs on the left)

What im having trouble with is the last expression on the LHS. I have (c-yb)^b so how am I supposed to solve for y, if i dont know b?
It's very difficult to read what you wrote (use LaTex!!), but there's an error here:

(1/c)ln|y| - (b/c)ln|c - yb| = t + K

That should be: ##\displaystyle \frac{1}{c}\ln|y| - \frac{1}{b}.\frac{b}{c}\ln|c-yb| = t + K## so that coefficient of b should cancel out.
 

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