How do you solve for dp/dt=kP(1-P/M)

1. Jul 20, 2009

brandy

1. The problem statement, all variables and given/known data
dy/dx=ky(1-y/M)
where p is smaller than m and k is positive

3. The attempt at a solution
∫dy/(ky(1-y/M))=x
∫dy/(ky(1-y/M))=∫(a/ky + b/(1-y/M))dy
then i introduced 2 new variables (a and b)
multiplied them out so that
1/(ky(1-y/M))=(a(1-y/M)+bky)/(ky(1-y/M))
cancelled out the lower half and you have
1=a(1-y/M)+bky

then i get a little stuck, im not even sure you can say that with the A and B becuase when you have 1/(x+2)^2 it equals a/(x+2)+b/(x+2)^2.
so yes. help?

Last edited: Jul 21, 2009
2. Jul 20, 2009

HallsofIvy

That should be "dx" on the right side, not just "x".

This is written confusingly- like the "x" magically turned into a/ky+ b/(1- y/m). Are "M" and "m" the same? Actually, this new equation does not derive from the previous one- you might want to write it separately or better, explain exactly what you are doing. You are writing the fraction on the left side in "partial fractions". 1/(ky(1- y/M))= a/ky+ b(1- y/M). Now multiply both sides of the equation by y(1- y/M): 1= a(1- y/M)+ bky.

Set y equal to things that make the original denominator 0: if y= 0, that becomes 1= a(1- 0)+ bk(0)= a. a=1. If y= M, 1= a(0)+ bkM so b= 1/M.

1/(ky(1-y/M))= 1/ky+ 1/M(1- y/M)= 1/ky+ 1/(M-y)

Now go back to your original equation: dy/ky+ dy/(M-y)= dx

Integate both sides.

then i get a little stuck, im not even sure you can say that.
so yes. help?[/QUOTE]

3. Jul 20, 2009

zcd

$$\frac{dP}{dt} = kP(1-\frac{P}{M})$$ is a separable differential equation, so you can separate it into $$\int \frac{dP}{P(M-P)} = \int \frac{k}{M}\,dt$$

4. Jul 20, 2009

AUMathTutor

Quite right, zcd. Then I believe the RHS is easy to evaluate, and the LHS needs a method of integration. Partial fraction decomposition, perhaps?

5. Jul 21, 2009

brandy

sorry. i just realised how many mistakes i made in typing that.
but im pretty sure i fixed them all now, so for clarification just read over the first post.

6. Jul 21, 2009

AUMathTutor

Well, it's still separable, and I still think partial fractions would work.

Anybody else?

7. Jul 21, 2009

zcd

Separating the variables early on will make it easier to solve. It seems you have the general idea here by attempting partial fraction decomposition
but you didn't quite follow through. You will get the equations $$\int\frac{1}{ky}+\frac{1}{k(y-M)}\,dy=x$$