# Find The General Solution [closed]

Using separation of variables and partial fractions find the general solution: $$dP/dt = re^{-at}p(1-p)$$

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## closed as not a real question by Andres Caicedo, Asaf Karagila, Ｊ. Ｍ., Zev ChonolesDec 25 '11 at 20:05

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I'll get right on that. –  The Chaz 2.0 Sep 27 '11 at 4:51
Which step did you have problems with? Did you manage to separate the variables? If so, did you manage to write both sides in an easily integrateable form? –  Jyrki Lahtonen Sep 27 '11 at 5:27
$\text{}$Ellecia, it'd be nice if you (a) didn't command us to do your problem - hence the downvotes and (not my) close vote - and (b) tell us what you've tried and where you might be stuck. (And possibly (c) tag this as [homework] if it is such.) Do you understand what separation of variables means? Do you see where partial fraction decomposition is useful? –  anon Sep 27 '11 at 5:34

"Separate" the variables. In the usual notation, we arrive at $$\frac{dp}{p(1-p)}=re^{-at}\,dt.$$ (I assume that the $P$ on the left of the posted equation is meant to be $p$.) There is a complication we will take care of later: Division by $p(1-p)$ makes no sense if $p=0$ or $p=1$. Now integrate.
The integral $\int re^{-at}\,dt$ is straightforward. We get $$\int re^{-at}\,dt=-\frac{r}{a}e^{-at}+C$$ if $a \ne 0$. If $a=0$ the integration is even easier. It is probably intended, though unsaid, that $a \ne 0$. So from now on we assume that $a \ne 0$.
Now we want $\int \frac{dp}{p(1-p)}$. We use partial fractions. It is not hard, either by a systematic method or by just fooling around, to find that $$\frac{1}{p(1-p)}=\frac{1}{p}+\frac{1}{1-p}.$$ Thus $$\int \frac{dp}{p(1-p)}=\ln(|p|)-\ln(|1-p|) +D$$ and we conclude that $$\ln\left(\left|\frac{p}{1-p}\right|\right)+D=-\frac{r}{a}e^{-at} +C.$$ In a fair number of problems, one really cannot go any further. But here we can solve explicitly for $p$, and we would almost certainly be expected to do so.
Collapse the two arbitrary constants of integration into one constant $E$ on the right. Then find the exponential of both sides. We get $$\left|\frac{p}{1-p}\right|=e^E e^{-\frac{r}{a}e^{-at}}=Ke^{-\frac{r}{a}e^{-at}}$$ where $K$ is a positive constant. Removing the absolute value signs yields $$\frac{p}{1-p}=\pm K e^{-\frac{r}{a}e^{-at}}.$$
There is a slight complication. The analysis has missed the solutions $p=0$ and $p=1$. These can be identified informally with the cases $K=0$ and $K=\infty$. Or else, maybe more responsibly, we note that $p=0$ and $p=1$ are solutions, and for the other ones we have $$\frac{p}{1-p}=Le^{-\frac{r}{a}e^{-at}}$$ for some non-zero constant $L$. The last equation can be rewritten as $p=(1-p)Le^{-\frac{r}{a}e^{-at}}$, and now we have a linear equation to solve for $p$.