ODE Population problem
I know that the rate of change with time of a population is proportional to the square root of t. T=0 is y = 100. Population increases at rate of 20 per month.
I started out by trying to do dy/dt = p^.5. I am used to the population problems where I use y=Ce^(rt) but am having trouble making the jump to this kind. How do you account for the increase of 20/month? 
Re: ODE Population problem
The population always increases at a rate of 20 per month, or is that an initial value?

Re: ODE Population problem
Yes, always increases at rate of 20/month and initial population is 100.

Re: ODE Population problem
Than the question makes no sense to me. How can the rate of increase in the population be of 20/month and proportional to the square root of t at the same time?

Re: ODE Population problem
You and me both then. Here is the exact question:
The time rate of change of a rabbit population P is proportional to the square root of P. At time t = 0 (months) the population numbers 100 rabbits and is increasing at the rate of 20 rabbits per month. How many after one year? The answer I guess is 484. How do you set up logistics equations? 
Re: ODE Population problem
So 20/month is an initial value.
The rate of change of P, i.e. it's derivative with respect to P is proportional to the square root of P: [tex] \frac{dP(t)}{dt} = \alpha P^{1/2} [/tex] where alpha is a constant with respect to. You also know that [tex] \frac{dP(t)}{dt} \bigg_{t=0}} = 20 \ rabbits/month [/tex] and that [tex]P(0) = 100 \ rabbits[/tex] Which is all the info you need to determine P(t) 
Re: ODE Population problem
I appreciate what you wrote but I'm still unsure on where to start. I see that I was wrong to have a p^.5 on the right. Should I go ahead and start separating variables and then moving forward?

Re: ODE Population problem
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Of course you should. Even in the case where you shouldn't, equations don't bite you know. 
Re: ODE Population problem
But how do you incorporate the dp/dt = 20 and P(0) = 100? I've done the sep of variables, no problem there. Sitting on 2p^.5 = (alpha)t + C.

Re: ODE Population problem
Made another attempt at this, can you tell me where I'm going wrong?
I came up with this equation after sep of variables: 2p^.5 = Bt + C. 1) With t = 0 and initial population of 100, I came up with a C value of 100. 2) With t = 1, population = 120, so I came up with a "B" value of 120. 3) But then when I do time = 12, I don't get anywhere near the 484 rabbits that I should have. 
Re: ODE Population problem
Er, did you solve for p in your equation so you have some p(t) on the left side instead of 2p^.5?

Re: ODE Population problem
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