Question

In Problems 1–22 Solve the given differential equation by separation of variables.$\frac{\mathbf{\text{dP}}}{\mathbf{\text{dt}}}{\mathbf{\text{=P-P}}}^{\mathbf{\text{2}}}$

Short answer

$P\left(t\right)=\frac{C{e}^t}{1+C{e}^t}$

Step-by-step solution

Step 1:Definition of separable equation

A first-order differential equation of the form$\frac{dy}{dx}=g\left(x\right)h\left(y\right)$is said to be separable or to have separable variables.

Step 2:Separate the variables and integration

$\frac{dP}{dt}=P-P^2$

Setup the integrals of the differential equation by means of separation of variables.

$\int \frac{dP}{P-P^2}=dt$

$p\left(P\right)\frac{dP}{dt}=g\left(t\right)$

$\int \frac{1}{P-P^2}dP=\int dt$

Solve the integral of$\int \frac{1}{P-P^2}dP$by separating the fraction into simpler parts by means of partial fraction decomposition

$\int \frac{1}{P-P^2}dP$

Then integrate by means of basic integration.

Step 3:Integrate by parts

$$\begin{gathered} \int \frac{1}{P(1-P)}dP \\ \begin{array}{c}\frac{1}{P(1-P)}=\frac{A}{P}+\frac{B}{1-P}\\ 1=A(1-P)+BP\\ BP-AP=0\to A-B=0\\ A=1\\ A=B\to B=1\end{array} \end{gathered}$$

$\begin{array}{l}\int \frac{1}{P}+\frac{1}{1-P}dP\\ \ln \left|P\right|-\ln |1-P|\\ \int \frac{1}{P-P^2}dP=\ln \left|\frac{P}{1-P}\right|\end{array}$

Solve the integral by means of basic integration.

$\int dt$

$\int dt=t+C$

Step 4:Equating the equations

Equate both equations and solve for the general solution of$P\left(t\right)$

$\begin{array}{c}\ln \left|\frac{P}{1-P}\right|=t+C\\ \frac{P}{1-P}=e^{t+C}\\ \frac{P}{1-P}=e^t{e}^C\\ \frac{P}{1-P}=C{e}^t\end{array}$

$\begin{array}{c}P=C(1-P)e^t\\ P=C{e}^t-CP{e}^t\\ P+CP{e}^t=C{e}^t\\ P(1+C{e}^t)=C{e}^t\\ P\left(t\right)=\frac{C{e}^t}{1+C{e}^t}\end{array}$

Hence the solution is$\mathbf{\text{P(t)=}}\frac{{\mathbf{\text{Ce}}}^{\mathbf{\text{t}}}}{{\mathbf{\text{1+Ce}}}^{\mathbf{\text{t}}}}$