Q. 76 - Calculus

Question

 Prove Theorem 6.21 by solving the initial-value problem $\frac{dP}{dt}=rP 1-\frac{P}{K}$with P(0) = P0, where r and K are constants

Step-by-Step Solution

Answer

Proved

Step 1. Given

The given initial-value problem is $\frac{dP}{dt}=rP 1-\frac{P}{K}$.

Step 2. Proof

Observe that the differential equation does not involve the independent variable at all, so solve the differential equation by antidifferentiation method

$\int \frac{dP}{P\left(1-{\displaystyle \frac{P}{K}}\right)}=rdt$

The Integrand on the left hand side needs to be simplified by the use of partial fractions. So, first resolve the fraction in to partial fractions by using cover up rule

$$\begin{gathered} \frac{1}{P\left(1-{\displaystyle \frac{P}{K}}\right)}=\frac{A}{P}+\frac{B}{1-{\displaystyle \frac{P}{K}}} \\ A=1 \\ B=\frac{1}{K} \\ H\mathrm{ence}, \mathrm{Integral} \mathrm{on} \mathrm{the} \mathrm{left} \mathrm{hand} \mathrm{side} \mathrm{comprises} \mathrm{of} \mathrm{two} \mathrm{integrals}. \mathrm{Solve} \mathrm{these} \mathrm{integrals} \mathrm{as} \\ \int \frac{1}{\mathrm{P}}\mathrm{dP}+\frac{1}{\mathrm{K}}\int \frac{\mathrm{dP}}{1-{\displaystyle \frac{\mathrm{P}}{\mathrm{K}}}}=\mathrm{r}\int \mathrm{dt} \\ \ln \left|\mathrm{p}\right|+\mathrm{k} \ln \left|1-\frac{\mathrm{P}}{\mathrm{K}}\right|=\mathrm{rt}+{\mathrm{C}}_1 \\ \mathrm{simplifying} \mathrm{the} \mathrm{equations} \\ \mathrm{P}=\frac{\mathrm{K}}{1+{\mathrm{Ae}}^{-\mathrm{rt}}} \\ \mathrm{Now} \mathrm{take} \mathrm{t}=0, \mathrm{P}={\mathrm{P}}_0 \mathrm{in} \mathrm{the} \mathrm{above} \mathrm{result} \mathrm{and} \mathrm{evaluate} \mathrm{the} \mathrm{constant} \mathrm{A} \\ {\mathrm{P}}_0=\frac{\mathrm{K}}{1+\mathrm{A}} \\ \mathrm{A}=\frac{\mathrm{K}-{\mathrm{P}}_0}{{\mathrm{P}}_0} \\ \mathrm{P}\left(\mathrm{t}\right)=\frac{{\mathrm{KP}}_0}{{\mathrm{P}}_0+(\mathrm{K}-{\mathrm{P}}_0){\mathrm{e}}^{-\mathrm{rt}}} \end{gathered}$$