I am currently trying this DE, and I am stuck on the question I stated below: \begin{align} \frac{dP}{dt}&=P-P^2 \\ \frac{dP}{P-P^2}&=dt \\ \int \frac{dP}{P(1-P)}&=\int dt \end{align}

From here I decompose the left using partial fraction decomposition.

Partial Decomposition:

\begin{align} \frac{A}{P}+\frac{B}{1-P}&=\frac{1}{P-P^2} \\ A(1-P)+B(P)&=1 \\ A-A(P)+B(P)&=1+ 0P \\ A+P(-A+B)&=1+0P \\ A&=1 \\-A+B&=0 \\ B&=1 \end{align}

$$\int \frac{1}P+\frac{1}{1-P} dP = \int dt$$ I get the following as the answer: $$\ln(\lvert P \rvert) - \ln(\lvert 1-P \rvert) = t + C$$ Why is the second term in the left hand side negative?


Because given $$\int \frac{1}P+\frac{1}{1-P} dP = \int\frac 1P \,dp+ \int \frac 1{1-P} \,dP,$$

when integrating $$\int \frac 1{1-P}\, dP$$ we have $\color{blue}{u = 1-P}$ so $\color{blue}{du= -dP}$, or $\color{blue} {dP = -\,du}.$ Hence $$\int \frac 1{1-P} \, dP = \int -\frac{du}{u} = - \ln|u| + c= -\ln|1-P|+ c$$


Because the coefficient of $P$ is $-1$.
So the integral becomes, $$\frac{\ln|1-P|}{-1} = -\ln|1-P|$$


You already have a couple of correct answers. I would encourage you to remember the following:

$ \int \frac{1}{ax+b}dx = \frac{1}{a}\ln|ax+b| +c. $ A lot of my DE students do this common mistakes. Just a matter of craefulness.

Another thing: I have seen many of my students tend to believe that $ \int \frac{1}{f(x)}dx =\ln|f(x)| +c$, which is absoutely incorret. I have warned them a couple of times. Still, they do the same mistakes. However,$ \int \frac{f^{\prime}(x)}{f(x)}dx = \ln|f(x)| +c$. This is correct. You can prove it easily. I hope this helps a bit.


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