Problems solving the Differential Equation

\frac{ dN }{ dt }=rN(1-N/K)

I first distributed the "N", found a comon denominator and separated the variables. "K" would be a function of "N" and "r" would be a function of "t"

Using Partial Fraction Decomposition, I ended up with a DE that I can solve.

The stipulation, and where I am having trouble is solving for the initial value condition.

N(0)=N_{0}

I will attach a image showing the work done.

My book has N(t)= \frac{N_{0}*Ke^{rt}}{K+N_{0}*(e^{rt}-1)}

Thanks

Question

Problems solving the Differential Equation

\frac{ dN }{ dt }=rN(1-N/K)

I first distributed the "N", found a comon denominator and separated the variables. "K" would be a function of "N" and "r" would be a function of "t"

Using Partial Fraction Decomposition, I ended up with a DE that I can solve.

The stipulation, and where I am having trouble is solving for the initial value condition.

N(0)=N_{0}

I will attach a image showing the work done.

My book has N(t)= \frac{N_{0}*Ke^{rt}}{K+N_{0}*(e^{rt}-1)}

Thanks

anonymous · Answer

Problems:

Problems solving the Differential Equation

\frac{ dN }{ dt }=rN(1-N/K)

I first distributed the "N", found a comon denominator and separated the variables. "K" would be a function of "N" and "r" would be a function of "t"

Using Partial Fraction Decomposition, I ended up with a DE that I can solve.

The stipulation, and where I am having trouble is solving for the initial value condition.

N(0)=N_{0}

I will attach a image showing the work done.

My book has N(t)= \frac{N_{0}*Ke^{rt}}{K+N_{0}*(e^{rt}-1)}

Thanks

anonymous · Answer

$$\frac{ dN }{ dt }=rN(1-N/K)$$

anonymous · Answer

$$N(0)=N_{0}$$

anonymous · Answer

$$N(t)= \frac{N_{0}*Ke^{rt}}{K+N_{0}*(e^{rt}-1)}$$

anonymous · Answer

u say "K" would be a function of "N" and "r" would be a function of "t" but u trated K and r like a constant

anonymous · Answer

*treated

anonymous · Answer

Yes, they are constants but with respect to either time or population

anonymous · Answer

This is a logistic model, one of which I am just learning. The solution in the book shows that "K" and "r" must be integrated on separate sides. I belive the work to be correct but can not for the life of me find how the solution is my 5th post

anonymous · Answer

u got$$\frac{N}{K-N}=ce^{rt}$$$$N(0)=N_0$$it gives$$c=\frac{N_0}{K-N_0}$$so u have$$\frac{N}{K-N}=\frac{N_0}{K-N_0}e^{rt}$$$$N+N\frac{N_0}{K-N_0}e^{rt}=K\frac{N_0}{K-N_0}e^{rt}$$$$N=\frac{K\frac{N_0}{K-N_0}e^{rt}}{1+\frac{N_0}{K-N_0}e^{rt}}$$multiply num and denum by $K-N_0$ u will get$$N=\frac{KN_0e^{rt}}{K-N_0+N_0e^{rt}}=\frac{N_{0}*Ke^{rt}}{K+N_{0}*(e^{rt}-1)}$$

anonymous · Answer

How do we turn N into $$N_{0}$$? Solving for $$N(0)=N_{0}$$ Wouldnt that mean t=0? Why would be assume/turn N into $$N_{0}$$

anonymous · Answer

well $N=N(t)$ is a function of time and $N(t=0)=N_0$ is a given condition of problem

anonymous · Answer

when we let $t=0$ then  $N$ turns to $N(t=0)=N_0$

anonymous · Answer

Man, I should of saw that. It is a function of t. So they will convert to the initial value. Thanks for the insight!!

anonymous · Answer

yw :)