Thread: I know you're gonna dig this Maths enthusiasts (challenging)

1. I know you're gonna dig this Maths enthusiasts (challenging)

Birth rates of animal populations are typically not constant but vary periodically with the passing seasons. Find an expression for P(t) if the population dynamics are governed by the differential equation:

dP/dt=(k+bcos2(pi)t)P

Where 't' is in years and k>0. Compare the growths of a population where b>0 with one where b=0 at the end of a full year. Both populations have the same initial value Po

2. Re: I know you're gonna dig this Maths enthusiasts (challenging)

Pretty straight forward separable equations.

With b> 0 you have $\displaystyle \frac{dP}{P}= (k+ b cos(2\pi t) dt$ and with b= 0, $\displaystyle \frac{dP}{P}= k dt$.

Those are easy to integrate.

3. Re: I know you're gonna dig this Maths enthusiasts (challenging)

Originally Posted by HallsofIvy
Pretty straight forward separable equations.

With b> 0 you have $\displaystyle \frac{dP}{P}= (k+ b cos(2\pi t) dt$ and with b= 0, $\displaystyle \frac{dP}{P}= k dt$.

Those are easy to integrate.
What is the answer after integration?

4. Re: I know you're gonna dig this Maths enthusiasts (challenging)

Originally Posted by NFS1
What is the answer after integration?
You can't integrate \displaystyle \displaystyle \begin{align*} \int{\frac{1}{P}\,dP} \end{align*} and \displaystyle \displaystyle \begin{align*} \int{k\,dt} \end{align*}?

5. Re: I know you're gonna dig this Maths enthusiasts (challenging)

Well, that's for KayAB to answer! They are pretty straight forward integrations. And since KayAB is asking about differential equations, I assume s/he can do integration.

6. Re: I know you're gonna dig this Maths enthusiasts (challenging)

Originally Posted by HallsofIvy
Well, that's for KayAB to answer! They are pretty straight forward integrations. And since KayAB is asking about differential equations, I assume s/he can do integration.
KayAB has gone AWOL so can u solve the mystery?

7. Re: I know you're gonna dig this Maths enthusiasts (challenging)

Originally Posted by NFS1
KayAB has gone AWOL so can u solve the mystery?
\displaystyle \displaystyle \begin{align*} \int{k\,dt} \end{align*}
k is a constant, so \displaystyle \displaystyle \begin{align*} \int{k\,dt} \end{align*} = \displaystyle \displaystyle \begin{align*} k \int{,dt} \end{align*}