Chapter 7.1, Problem 48E

a.

To determine

To graph:

To graph the given initial value problem $\frac{dy}{dx}=-x^{-2}$ and $y=1$ when $x=1$

Answer to Problem 48E

  

Explanation of Solution

Given information:

The given initial value problem is $\frac{dy}{dx}=-x^{-2}$ and $y=1$ when $x=1$

Graph:

  

Now to find the general solution by finding the anti-derivative.

  $\begin{array}{l}\frac{dy}{dx}=-x^{-2}\\ y={\displaystyle \int \left(-x^{-2}\right)dx}\\ y=x^{-1}+c\end{array}$ [Where c is constant]

Now substituting the initial condition of $y=1$ when $x=1$ into the general solution and solve for c.

  $\begin{array}{l}1=1^{-1}+c\\ 1=1+c\Rightarrow c=0\end{array}$

Substitute the value of c into the general solution to find the particular solution of the differential equation.

  $\begin{array}{l}y=x^{-1}+0\\ y=x^{-1}\end{array}$

So the solution of the initial value problem is

  $y=x^{-1}$

Interpretation:

When a curve is discontinuous, the initial condition only gives us the continuous piece of the curve that passes through the given point.

Therefore, when the graph $y=x^{-1}$ ,only graph the right half of the point through that passes through $\left(1,1\right)$ .

b.

To determine

To Show:

To show that the given graph is not the correct answer of part (a).

Answer to Problem 48E

The given graph is an incorrect graph to part (a).

Explanation of Solution

Given information:

The given graph is

  

Proof:

When a curve is discontinuous, the initial condition only gives you the part of the graph that passes through that point.

The curve has a discontinuity at $x=0$ so the point $\left(1,1\right)$ only gives the portion of the curve to the right of $x=0$ since $\left(1,1\right)$ lies to the right of $x=0$ .

The graph is incorrect because it also includes the portion of the graph to the left of $x=0$ .

Hence, the given graph is not correct to part (a).