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

[brandy]

Homework Statement

dy/dx=ky(1-y/M)
where p is smaller than m and k is positive

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, I am not even sure you can say that with the A and B because when you have 1/(x+2)^2 it equals a/(x+2)+b/(x+2)^2.
so yes. help?

 

[HallsofIvy]

Homework Statement

dy/dx=ky(1-y/M)
where p is smaller than m and k is positive

The Attempt at a Solution

dy/(ky(1-y/M))=x

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

dy/(ky(1-y/M))=a/ky + b/(1-y/m)
1=a(1-y/m)+bky

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, I am not even sure you can say that.
so yes. help?[/QUOTE]

 

[zcd]

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

 

[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?

 

[brandy]

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

 

[AUMathTutor]

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

Anybody else?

 

[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

∫dy/(ky(1-y/M))=x
∫dy/(ky(1-y/M))=∫(a/ky + b/(1-y/M))dy

but you didn't quite follow through. You will get the equations $$\int\frac{1}{ky}+\frac{1}{k(y-M)}\,dy=x$$