Yarr Need help with a differential equation

[CaptainJames]

Can't get this one to work right. 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 rabits will there be one year later?

 

[CaptainJames]

Oops! Didn't show what I've done. Okay here goes... well first you know that at time t=0 the population is 100 and the rate of change is 20, so k =2.
dP/dt = 2 times square root of p --(is there any way for mathmatical symbols?) so... dP/(k sqaure root of p) = dt

2k square root of p = t + c

square root of p = (t+c)/(2k)

p = ((t+c)/(2k))^2

100 = ((0+c)/4))^2
c=40.
At this point when I try and solve it t isn't in terms of months it's in something i don't know. Anyone know what I did wrong?

Sorry, this is barely readable.. this is my first post, so I'm going to touch it up a bit

 

[xman]

your solution method is fine, but I'm getting
$$\frac{dP}{\sqrt{P}}=kdt \Rightarrow P(t) = \frac{k^{2}}{4} \left(t+const\right)^{2}$$
from which you may apply the initial conditions to get the constants. hope this helps, sincerely, x

also do you know LaTeX/ TeX type setting? if so click on the image and see how the the text is made, say for this equation above.

 

[CaptainJames]

Thanks so much! What did I do wrong...?

 

[xman]

Thanks so much! What did I do wrong...?

look at your first step. you have the problem set up, but... do you see it?

 

[CaptainJames]

Oh crap, yea I see it. I said dP/(2k square root of p) = 2k square root of P, instead of 1/2k * square root of P. Thanks for the help :P.

 

[xman]

no problem, good work, do you understand how to type set click on my post with the differential equation and note the use of the TeX in the boxes to write equations. in fact if you go to just about any post with equations written like the one above then, if you don't know TeX, you can learn...if you want a way to post formulas and all that is.