Chapter 7.3, Problem 24E

To determine

To solve : the differential equation and graph several members of the family of solutions.

Answer to Problem 24E

The solution of differential equation is $y=\ln |\sin x+c|$ .

Explanation of Solution

Given information :

The differential equation is $e^{-y}y\text{'}+\cos x=0$ .

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

  $\begin{array}{l}{e}^{-y}y\text{'}+\cos x=0\\ e^{-y}\frac{dy}{dx}+\cos x=0\text{ [}y\text{'}=\frac{dy}{dx}]\\ e^{-y}dy=-\cos xdx\text{ [separate the variables]}\\ {\displaystyle \int e^{-y}dy=}{\displaystyle \int -\cos xdx\text{ [integrate both side]}}\\ -e^{-y}=-\sin x+c\text{ [simplify integrate] }\\ e^y\text{=}\sin x+c\text{ [multiply by -1]}\\ y=\ln |\sin x+c|\text{ [take ln both side]}\end{array}$

Here $c$ is a arbitrary constant.

Hence, the solution of differential equation is $y=\ln |\sin x+c|$ .

Now substitute different value for c to get different members of family of solutions.

  $\begin{array}{l}y=\ln |\sin x+1|\text{ [}c=1]\\ y=\ln |\sin x+2|\text{ [}c=2]\\ y=\ln |\sin x+3|\text{ [}c=3]\\ y=\ln |\sin x-2|\text{ [}c=-2]\\ y=\ln |\sin x-1|\text{ [}c=-1]\end{array}$

Using implicit plotting capability of a CAS to graph the curves $y=\ln |\sin x+1|,y=\ln |\sin x+2|,y=\ln |\sin x+3|,y=\ln |\sin x-2|,y=\ln |\sin x-1|$ .

The graph is obtained as:

  

When constant c value is not $1\text{ or }-1$ , the curve of graph is above from the x -axis and for $1\text{ or }-1$ the curve of graph mainly downwards of x -axis but have real roots.