1/P dP/dt=b+at

davelml · Jan 13, 2011, 10:42 AM

at t=o P(0)=3.9 I am not sure how to work this problem. I have seen different solutions for this problem and some of them do not work for accurately predicting P(t). This is for populations.

Unknown008 · Jan 13, 2011, 12:09 PM

Your question is not clear at all...

Can you post the original question, as you were given?

galactus · Jan 13, 2011, 01:04 PM

Assuming you mean to solve:

$\frac{1}{P}\cdot \frac{dP}{dt}=b+at$ when P(0)=3.9

you can solve the equation per the initial condition and get:

$P=3.9e^{\frac{at^{2}}{2}+bt}$

ebaines · Jan 13, 2011, 02:25 PM

I get:

$ P = 3.9 e^{(bt + \frac {at^2} 2)} \ =\ 3.9 e^{bt} e^{\frac {a t^2} 2} $

Like this:

$ \frac {dP} {dt} = b+at \\ \\ \frac {dP} P = (b + at)dt \\ \\ \int \frac {dP} P = \int (b + at)dt \\ \\ ln(P) = bt + \frac 1 2 a t^2 + c \\ \\ P = e^{(bt + \frac 1 2 a t^2 + c)} \ = \ e^{(bt + \frac 1 2 a t^2 )} e^c $

From the given boundary condition you know that $e^c = 3.9$

galactus · Jan 13, 2011, 03:17 PM

Yep, ebaines. That is what I got as well. I do not know why I typed what I initially typed.
Brain fart, I suppose.