Chapter 7.1, Problem 75E

a.

To determine

To find: A solution to the differential equation in the interval $\left(0,\infty \right)$ that satisfies $y\left(1\right)=2$ .

Answer to Problem 75E

A solution to the differential equation

  $y=\frac{x^2}{2}+\frac{1}{x}+\frac{1}{2}$

Explanation of Solution

Given information: Differential equation

  $\frac{dy}{dx}=x-\frac{1}{x^2}$

Calculation:

Differential equation

  $\begin{array}{l}\frac{dy}{dx}=x-\frac{1}{x^2}\\ {\displaystyle \int dy}={\displaystyle \int \left(x-\frac{1}{x^2}\right)}dx\\ y={\displaystyle \int xdx-{\displaystyle \int \frac{1}{x^2}}}dx\\ y=\frac{x^2}{2}+\frac{1}{x}+c\\ \text{Given }y\left(1\right)=2\\ c=\frac{1}{2}\\ y=\frac{x^2}{2}+\frac{1}{x}+\frac{1}{2}\end{array}$

b.

To determine

To find: A solution to the differential equation in the interval $\left(-\infty ,0\right)$ that satisfies $y\left(-1\right)=1$ .

Answer to Problem 75E

solution to the differential equation

  $y=\frac{x^2}{2}+\frac{1}{x}+\frac{3}{2}$

Explanation of Solution

Given information: Differential equation

  $\frac{dy}{dx}=x-\frac{1}{x^2}$

Calculation:

Differential equation

  $\begin{array}{l}\frac{dy}{dx}=x-\frac{1}{x^2}\\ {\displaystyle \int dy}={\displaystyle \int \left(x-\frac{1}{x^2}\right)}dx\\ y={\displaystyle \int xdx-{\displaystyle \int \frac{1}{x^2}}}dx\\ y=\frac{x^2}{2}+\frac{1}{x}+c\\ \text{Given }y\left(-1\right)=1\\ c=\frac{3}{2}\\ y=\frac{x^2}{2}+\frac{1}{x}+\frac{3}{2}\end{array}$

c.

To determine

To prove: That the following piecewise function is a solution to the differential equation for any values of $C_1$ and $C_2$ .

Explanation of Solution

Given information: Piecewise function

  $y=\left\{\begin{array}{l}\frac{x^2}{2}+\frac{1}{x}+C_1\text{ x<0}\\ \frac{x^2}{2}+\frac{1}{x}+C_2\text{ x>0}\end{array}\right\}$

Proof:

Piecewise function

  $\begin{array}{l}y=\left\{\begin{array}{l}\frac{x^2}{2}+\frac{1}{x}+C_1\text{ x<0}\\ \frac{x^2}{2}+\frac{1}{x}+C_2\text{ x>0}\end{array}\right\}\\ \frac{dy}{dx}=\left\{\begin{array}{l}x-\frac{1}{x^2}\text{ x<0}\\ x-\frac{1}{x^2}\text{ x>0}\end{array}\right\}\\ \text{Therefore }\frac{dy}{dx}=x-\frac{1}{x^2}\end{array}$

So, the piecewise function is a solution of the differential equation.

  $\frac{dy}{dx}=x-\frac{1}{x^2}$

d.

To determine

To calculate: Values for $C_1$ and $C_2$ .

Answer to Problem 75E

Values for $C_1\text{=}\frac{1}{2}$ and $C_2\text{=}\frac{3}{2}$ .

Explanation of Solution

Given information: Piecewise function

  $y=\left\{\begin{array}{l}\frac{x^2}{2}+\frac{1}{x}+C_1\text{ x<0}\\ \frac{x^2}{2}+\frac{1}{x}+C_2\text{ x>0}\end{array}\right\}$

Calculation:

A solution to the differential equation in the interval $\left(0,\infty \right)$ that satisfies $y\left(1\right)=2$

is $y=\frac{x^2}{2}+\frac{1}{x}+\frac{1}{2}$ .

A solution to the differential equation in the interval $\left(-\infty ,0\right)$ that satisfies $y\left(-1\right)=1$

is $y=\frac{x^2}{2}+\frac{1}{x}+\frac{3}{2}$ .

Values for $C_1\text{=}\frac{1}{2}$ and $C_2\text{=}\frac{3}{2}$

e.

To determine

To calculate: Values for $C_1$ and $C_2$ .

Answer to Problem 75E

Values for $C_1\text{=}\frac{1}{2}$ and $C_2=\frac{-7}{2}$ .

Explanation of Solution

Given information: Piecewise function

  $y=\left\{\begin{array}{l}\frac{x^2}{2}+\frac{1}{x}+C_1\text{ x<0}\\ \frac{x^2}{2}+\frac{1}{x}+C_2\text{ x>0}\end{array}\right\}$

And $y\left(2\right)=-1,y\left(-2\right)=2$

Calculation:

Piecewise function

  $y=\left\{\begin{array}{l}\frac{x^2}{2}+\frac{1}{x}+C_1\text{ x<0}\\ \frac{x^2}{2}+\frac{1}{x}+C_2\text{ x>0}\end{array}\right\}$

And $y\left(2\right)=-1$

  $\begin{array}{l}y=\frac{x^2}{2}+\frac{1}{x}+C_2\\ y\left(2\right)=-1\\ C_2=\frac{-7}{2}\end{array}$

For $y\left(-2\right)=2$

  $\begin{array}{l}y=\frac{x^2}{2}+\frac{1}{x}+C_1\\ y\left(-2\right)=2\\ C_1=\frac{1}{2}\end{array}$

Values for $C_1\text{=}\frac{1}{2}$ and $C_2=\frac{-7}{2}$