# Logistic growth model, differential equation

• Wiz14

## Homework Statement

dY/dt = y(c - yb)

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

## 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 I am 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 don't know b?

## Homework Statement

dY/dt = y(c - yb)

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

## 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 I am 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 don't 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.