Chapter 7.3, Problem 25E

To determine

To solve : the differential equation and graph the solution.

Answer to Problem 25E

The solution of differential equation is $\cos y\text{=}\cos x-1$ .

Explanation of Solution

Given information :

The differential equation is $y\text{'}=\frac{\sin x}{\sin y}$ .

  $y\left(0\right)=\frac{\pi }{2}$

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

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

Here $c$ is a arbitrary constant.

Since. $y\left(0\right)=\frac{\pi }{2}$ so substitute $x=0,y=\frac{\pi }{2}$ ,

  $\begin{array}{l}\cos y\text{=}\cos x+c\\ \cos \frac{\pi }{2}=\cos 0+c\text{ [substitute }x=0,y=\frac{\pi }{2}]\\ 0=1+c\text{ [}\cos \frac{\pi }{2}=0,\cos 0=1]\\ c=-1\text{ [subtract 1 from both side] }\end{array}$

Now substitute $c=-1$ in the solution,

  $\begin{array}{l}\cos y\text{=}\cos x+c\\ \cos y\text{=}\cos x-1\end{array}$

Hence the solution is $\cos y\text{=}\cos x-1$ .

Using implicit plotting capability of a CAS to graph the curves $\cos y\text{=}\cos x-1$ .

The graph is obtained as: