Chapter 7.1, Problem 76E

a.

To determine

To find: A solution to the differential equation in the interval $\left(0,\infty \right)$ .

Answer to Problem 76E

A solution to the differential equation in the interval $\left(0,\infty \right)$

  $y=\ln x+C$ .

Explanation of Solution

Given information: Differential equation

  $\frac{dy}{dx}=\frac{1}{x}$

Calculation:

Differential equation

  $\begin{array}{l}\frac{dy}{dx}=\frac{1}{x}\\ {\displaystyle \int dy}={\displaystyle \int \left(\frac{1}{x}\right)}dx\\ y=\ln \left|x\right|+C\end{array}$

For the interval $\left(0,\infty \right)$ $,\left|x\right|=x$

  $y=\ln x+C$

b.

To determine

To find: A solution to the differential equation in the interval $\left(-\infty ,0\right)$ .

Answer to Problem 76E

solution to the differential equation in the interval $\left(-\infty ,0\right)$

  $y=\ln (-x)+C$

Explanation of Solution

Given information: Differential equation

  $\frac{dy}{dx}=\frac{1}{x}$

Calculation:

Differential equation

  $\begin{array}{l}\frac{dy}{dx}=\frac{1}{x}\\ {\displaystyle \int dy}={\displaystyle \int \left(\frac{1}{x}\right)}dx\\ y=\ln \left|x\right|+C\end{array}$

For the interval $\left(-\infty ,0\right)$ $,\left|x\right|=-x$

  $y=\ln (-x)+C$

c.

To determine

To prove:Function $y=\ln \left|x\right|+C$ is a solution to the differential equation in the domain $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

Explanation of Solution

Given information: A solution to the differential equation in the interval $\left(0,\infty \right)$

  $y=\ln x+C$

And

solution to the differential equation in the interval $\left(-\infty ,0\right)$

  $y=\ln (-x)+C$

Proof:

A solution to the differential equation in the interval $\left(0,\infty \right)$

  $y=\ln x+C$

And

solution to the differential equation in the interval $\left(-\infty ,0\right)$

  $y=\ln (-x)+C$

  $\begin{array}{l}y=\left\{\begin{array}{l}y=\ln (-x)+C\text{ x<0}\\ y=\ln x+C\text{ x>0}\end{array}\right\}\\ \frac{dy}{dx}=\left\{\begin{array}{l}\frac{1}{x}\text{ x<0}\\ \frac{1}{x}\text{ x>0}\end{array}\right\}\\ \text{Therefore }\frac{dy}{dx}=\frac{1}{x}\text{ in the domain }\left(-\infty ,0\right)\cup \left(0,\infty \right)\end{array}$

d.

To determine

To prove: That the function

  $y=\left\{\begin{array}{l}y=\ln (-x)+C\text{ x<0}\\ y=\ln x+C\text{ x>0}\end{array}\right\}$

is a solution to the differential equation for any values of $C_1$ and $C_2$ .

Explanation of Solution

Given information: the function

  $y=\left\{\begin{array}{l}y=\ln (-x)+C_1\text{ x<0}\\ y=\ln x+C_2\text{ x>0}\end{array}\right\}$

Proof:

the function

  $\begin{array}{l}y=\left\{\begin{array}{l}y=\ln (-x)+C_1\text{ x<0}\\ y=\ln x+C_2\text{ x>0}\end{array}\right\}\\ \frac{dy}{dx}=\left\{\begin{array}{l}\frac{1}{x}\text{ x<0}\\ \frac{1}{x}\text{ x>0}\end{array}\right\}\end{array}$

Therefore, the piecewise is a solution to the differential equation for any values of $C_1$ and $C_2$ .