Chapter 7.3, Problem 20E

To determine

To find : the function.

Answer to Problem 20E

The equation of the curve is $y\left(1-y\right)=\frac{1}{4}{e}^x$ .

Explanation of Solution

Given information :

  $f\text{'}\left(x\right)=f(x)\left(1-f\left(x\right)\right)$

  $f\left(0\right)=\frac{1}{2}$

Calculation : to find the function, first solve the differential equation,

here the equation is separable, so write the equation in the terms of differentials and then integrate both sides,

let $y=f\left(x\right)$ ,

  $\begin{array}{l}f\text{'}\left(x\right)=f(x)\left(1-f\left(x\right)\right)\\ \frac{dy}{dx}=y\left(1-y\right)\text{ [}f\text{'}\left(x\right)=\frac{dy}{dx},y=f\left(x\right)]\\ \frac{dy}{y\left(1-y\right)}\text{=}dx\text{ [separate the variables]}\\ {\displaystyle \int \frac{dy}{y\left(1-y\right)}=}{\displaystyle \int dx\text{ [integrate both side]}}\\ \text{ln}|1-y|+\ln \left|y\right|\text{=}x+c\text{ [simplify integrate]}\\ \text{ln|}y\left(1-y\right)\text{|}=x+c\text{ [simplify]}\\ y\left(1-y\right)=e^{x+c}\text{ [remove ln]}\\ y\left(1-y\right)=A{e}^x\text{ [}{e}^{{}^c}=A]\end{array}$

HereA , $c$ is a arbitrary constant.

Since $f\left(0\right)=\frac{1}{2}$ ,

  $\begin{array}{l}y\left(1-y\right)=A{e}^x\\ \frac{1}{2}\left(1-\frac{1}{2}\right)=A{e}^0\text{ [substitute]}\\ \frac{1}{2}\left(\frac{1}{2}\right)=A\text{ [subtract]}\\ A=\frac{1}{4}\text{ [multiply]}\end{array}$

So the equation is $y\left(1-y\right)=\frac{1}{4}{e}^x$

Hence, theequation of the curve is $y\left(1-y\right)=\frac{1}{4}{e}^x$ .