Chapter 6.5, Problem 37E

To determine

To find: The expression $W$ as a function of $r$ under the constraint that $\text{W(0)=24}$ .

Answer to Problem 37E

The answer is $W=\frac{216}{1+8e^{-0.1728t}}$

Explanation of Solution

Given information:

The differential equation is $dW/dt=0.0008W(216-W)$ .

Calculation:

  $\begin{array}{c}\frac{1}{W(216-W)}dW=0.0008dt\\ {{\displaystyle \int }}^{\text{}}\frac{1}{W(216-W)}dW={{\displaystyle \int }}^{\text{}}0.0008dt\\ \frac{1}{W(216-W)}=\frac{A}{W}+\frac{B}{216-W}\\ =\frac{A(216-W)}{W(216-W)}+\frac{BW}{(216-W)W}\\ =\frac{A(216-W)+BW}{W(216-W)}\\ A(216-W)+BW=\mathrm{1.....}(1)\end{array}$

Put

  $W=216$ into the equation $(1)$ ,

And get,

  $\begin{array}{c}A(216-216)+B\cdot 216=1\\ \to A\cdot 0+216B=1\\ \to 216B=1\\ \to B=\frac{1}{216}\end{array}$

Put,

  $W=0$ into the equation $(1)$ ,

And get,

  $\begin{array}{c}A(216-0)+B\cdot 0=1\\ \to 216A+B.0=1\\ \to 216A=1\\ \to A=\frac{1}{216}\end{array}$

Therefore,

  $\frac{1}{W(216-W)}=\frac{1}{216W}+\frac{1}{W(216-W)}$

Thus,

  $\begin{array}{c}{{\displaystyle \int }}^{\text{}}\frac{1}{W(216-W)}dW={{\displaystyle \int }}^{\text{}}0.0008dt\\ {{\displaystyle \to \int }}^{\text{}}\left(\frac{1}{216W}+\frac{1}{216(216-W)}\right)dW={{\displaystyle \int }}^{\text{}}0.0008dt\\ {{\displaystyle \to \int }}^{\text{}}\frac{1}{216W}dW+{{\displaystyle \int }}^{\text{}}\frac{1}{216(216-W)}dW={{\displaystyle \int }}^{\text{}}0.0008dt\\ \to \frac{1}{216}\ln \left|W\right|-\frac{1}{216}\ln |216-W|=0.0008t+C\\ \to \ln |216-W|-\ln \left|W\right|=-0.1728t-C\\ \to \ln \frac{|216-W|}{\left|W\right|}=-0.1728t-C\\ \to \frac{216}{W}-1=e^{-0.1728t}\cdot e^{-C}\\ \to \frac{216}{W}=1+e^{-0.1728t}\cdot e^{-C}\end{array}$

Put,

  $\begin{array}{c}W=24\\ t=0\end{array}$

To find $C$ :

  $\begin{array}{c}\frac{216}{24}=1+e^{-0.1728\cdot 0}\cdot e^{-C}\\ \to 9=1+e^0\cdot e^{-C}\\ \to 8=e^{-C}\end{array}$

Thus,

  $\begin{array}{c}\frac{216}{W}=1+8e^{-0.1728t}\\ \to W=\frac{216}{1+8e^{-0.1728t}}\end{array}$

Therefore, the required expression $W$ as a function of $r$ under the constraint that $\text{W(0)=24}$ is:

  $W=\frac{216}{1+8e^{-0.1728t}}$