Ask Me Help Desk - 1/P dP/dt=b+at

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1/P dP/dt=b+at

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.

Your question is not clear at all...

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

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:

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

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 $e^c = 3.9$

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

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