# Trouble with Solving a Differential Equation

1. Feb 5, 2006

### xxdrossxx

I'm having trouble with the following textbook problem:

The only basic approach for solving differential equations that I know is to "separate and integrate", that is, have one side with P's and a dP, and the other side with t's and a dt (constants anywhere throughout), and then integrate both sides. So, what I was hoping to get was kM * something * dt = kP * something * dP. However, it looks like that's not going to happen. Could someone help me solve this thing?

I know that if "M is the maximum level of performance", then logically $$\lim_{t \rightarrow \infty} P(t) = M$$. However, I'm still not sure how to actually find the function P(t).

Last edited: Feb 5, 2006
2. Feb 5, 2006

### d_leet

It's easily separable just divide by by M-P and you have a pretty standard integral to find the function P.

3. Feb 5, 2006

### xxdrossxx

Ah, of course! I guess my first mistake was distributing the k. Dividing brings me to $$\frac{dP}{M - P} = kdt$$, and integrating both sides gives me $$-ln|M - P| = kt + c$$. Going from there:

$$ln|M - P| = -kt - c$$
$$e^{-kt - c} = M - P$$
$$P = M - e^{-kt - c}$$
I think I'm doing something very wrong here. The answer in the back of the book is $$P(t) = M - Me^{-kt}$$.

I'm thinking I've made a really stupid mistake, but I can't find it. Can anyone help?

Last edited: Feb 5, 2006
4. Feb 6, 2006

### HallsofIvy

Staff Emeritus
Have you left out part of the problem- like, say, the information that the subject knew nothing to start with? That is, that P(0)= 0.

$$P(t) = M - e^{-kt - c}= M- (e^{-c})e^{-kt}$$
and we can write that as
$$P(t)= M- C e^{-kt}$$
where $C= e^{-c}$.

Then, if P(0)= 0,
$$P(0)= M- Ce^0= M- C= 0$$
so C= M.

Actually that's not relevant to the final question: what is the limit of P(t) as t goes to infinity? That answer does not depend on C.