Chapter 6.1, Problem 47E

(a)

To determine

To sketch: The graph of the solution to the given initial value problem.

Answer to Problem 47E

The graph is:

  

Explanation of Solution

Given information:

The differential equation is: $\frac{dy}{dx}={\sec }^{2}x$

And $y=1$ when

  $x=\pi$

Calculation:

by resolving its anti derivative, one must locate the differential equation's general solution. At this time,

  $\begin{array}{c}f(x)=\int {\sec }^{2}xdx\\ =\tan x+C\end{array}$

Then apply the initial condition of $y=1$ when

  $x=\pi$ and get:

  $\begin{array}{c}f(\pi )=\tan (\pi )+C\\ =1\\ \Rightarrow 0+C=1\\ \Rightarrow C=1.\end{array}$

Now plug-in this value of $C$ to the general solution solved earlier. Therefore, the solution of the initial value problem is

  $y=\tan x+1$

The obtained solution's graph is then sketched:

  

The graph of $y=\tan x+1$ that passes through the point $(\pi ,1)$ .

(b)

To determine

To explain The given graph is not the correct answer to part $(a)$ .

Answer to Problem 47E

The graphing calculator used by the learner must only be used to plot one period of the graph.

Explanation of Solution

Given information:

The graph is:

  

Explain:

Since only interested in the graph of the solution that passes through the point $(\pi ,1)$ , claim that the student's graph is erroneous. The graphing calculator used by the learner must only be used to plot one period of the graph.