Chapter 6.1, Problem 75E

a.

To determine

Solving Differential Equations Let $\frac{dy}{dx}=x-\frac{1}{x^2}$ .

(a) Find a solution to the differential equation in the interval $(0,\infty )$ that satisfies $y(1)=2$ .

(b) Find a solution to the differential equation in the interval $(-\infty ,0)$ that satisfies $y(-1)=1$ .

(c) Show that the following piecewise function is a solution to the differential equation for any values of $C_1$ and $C_2$

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

(d) Choose values for $C_1$ and $C_2$ so that the solution in part (c) agrees with the solutions in parts (a) and (b).

(e) Choose values for $C_1$ and $C_2$ so that the solution in pan (c) satisfies $y(2)=-1$ and $y(-2)=2$ .

Find a solution to the differential equation in the interval $(0,\infty )$ that satisfies $y(1)=2.$

Answer to Problem 75E

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

Explanation of Solution

Given:

Differential equation: $\frac{dy}{dx}=x-\frac{1}{x^2}$ .

Condition: $y(1)=2$

Concept Used:

Variable separable method:

Calculation:

Differential equation: $\frac{dy}{dx}=x-\frac{1}{x^2}$

Apply variable separable method,

  $\begin{array}{l}\frac{dy}{dx}=x-\frac{1}{x^2}\\ dy=\left(x-\frac{1}{x^2}\right)dx\end{array}$

Integrate it,

  $\begin{array}{c}{\displaystyle \int dy}=\left({\displaystyle \int \left(x-\frac{1}{x^2}\right)dx}\right)+C\\ \text{Here, }C\text{ is arbitrary constant}\text{.}\\ \\ {\displaystyle \int dy}=\left({\displaystyle \int xdx}-{\displaystyle \int \frac{1}{x^2}dx}\right)+C\\ \\ {\displaystyle \int dy}=\left({\displaystyle \int xdx}-{\displaystyle \int x^{-2}dx}\right)+C\\ y=\frac{x^{1+1}}{1+1}-\frac{x^{-2+1}}{\left(-2+1\right)}+C\\ y=\frac{x^2}{2}-\frac{x^{-1}}{(-1)}+C\\ y=\frac{x^2}{2}+\frac{1}{x}+C\\ y=\frac{1}{x}+\frac{x^2}{2}+C\end{array}$

Condition: $y(1)=2$

Put $x=1,y=2$ in equation $y(x)=\frac{1}{x}+\frac{x^2}{2}+C$ .

  $\begin{array}{c}y(1)=\frac{1}{1}+\frac{1^2}{2}+C\\ 2=1+\frac{1}{2}+C\\ 2=\frac{3}{2}+C\\ C=2-\frac{3}{2}\\ C=\frac{1}{2}\end{array}$

Hence, solution is $y(x)=\frac{1}{x}+\frac{x^2}{2}+\frac{1}{2}$ .

Conclusion:

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

b.

To determine

Find a solution to the differential equation in the interval $(-\infty ,0)$ that satisfies $y(-1)=1.$

Answer to Problem 75E

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

Explanation of Solution

Given:

Differential equation: $\frac{dy}{dx}=x-\frac{1}{x^2}$

Condition: $y(-1)=1.$

Concept Used:

Variable separable method:

Calculation:

Differential equation: $\frac{dy}{dx}=x-\frac{1}{x^2}$

Apply variable separable method,

  $\begin{array}{c}\frac{dy}{dx}=x-\frac{1}{x^2}\\ dy=\left(x-\frac{1}{x^2}\right)dx\end{array}$

Integrate it,

  $\begin{array}{c}{\displaystyle \int dy}=\left({\displaystyle \int \left(x-\frac{1}{x^2}\right)dx}\right)+C\\ \text{Here, }C\text{ is arbitrary constant}\text{.}\\ \\ {\displaystyle \int dy}=\left({\displaystyle \int xdx}-{\displaystyle \int \frac{1}{x^2}dx}\right)+C\\ \\ {\displaystyle \int dy}=\left({\displaystyle \int xdx}-{\displaystyle \int x^{-2}dx}\right)+C\\ y=\frac{x^{1+1}}{1+1}-\frac{x^{-2+1}}{\left(-2+1\right)}+C\\ y=\frac{x^2}{2}-\frac{x^{-1}}{(-1)}+C\\ y=\frac{x^2}{2}+\frac{1}{x}+C\\ y=\frac{1}{x}+\frac{x^2}{2}+C\end{array}$

Condition: $y(-1)=1$

Put $x=-1,y=1$ in equation $y(x)=\frac{1}{x}+\frac{x^2}{2}+C$ .

  $\begin{array}{c}y(-1)=\frac{1}{\left(-1\right)}+\frac{{\left(-1\right)}^2}{2}+C\\ 1=-1+\frac{1}{2}+C\\ 1=\frac{-1}{2}+C\\ C=1+\frac{1}{2}\\ C=\frac{3}{2}\end{array}$

Hence, solution is $y(x)=\frac{1}{x}+\frac{x^2}{2}+\frac{3}{2}$ .

Conclusion:

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

c.

To determine

Show that the piecewise function is a solution to the differential equation for any values of $C_1$ and $C_2$

Piecewise function: $y=\left\{\begin{array}{l}\frac{1}{x}+\frac{x^2}{2}+C_{1}x<0\\ \frac{1}{x}+\frac{x^2}{2}+C_{2}x>0\end{array}\right.$

Answer to Problem 75E

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

Explanation of Solution

Given:

Differential equation: $\frac{dy}{dx}=x-\frac{1}{x^2}$

Piecewise function: $y=\left\{\begin{array}{l}\frac{1}{x}+\frac{x^2}{2}+C_{1}x<0\\ \frac{1}{x}+\frac{x^2}{2}+C_{2}x>0\end{array}\right.$

Calculation:

Differential equation: $\frac{dy}{dx}=x-\frac{1}{x^2}$

Case 1: when $x<0$ then

Solution is $y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_1$ .

If $y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_1$ is the solution of differential equation then it will satisfy the differential equation.

  $\begin{array}{c}\frac{dy}{dx}=x-\frac{1}{x^2}\\ \frac{d}{dx}\left(y\right)=x-\frac{1}{x^2}\\ \frac{d}{dx}\left(\frac{1}{x}+\frac{x^2}{2}+C_1\right)=x-\frac{1}{x^2}\left(\text{since }y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_1\right)\\ \frac{d}{dx}\left(\frac{1}{x}\right)+\frac{d}{dx}\left(\frac{x^2}{2}\right)+\frac{d}{dx}\left(C_1\right)=x-\frac{1}{x^2}\\ \frac{d}{dx}\left(x^{-1}\right)+\frac{d}{dx}\left(\frac{x^2}{2}\right)+\frac{d}{dx}\left(C_1\right)=x-\frac{1}{x^2}\\ -1x^{-2}+\left(\frac{2x}{2}\right)+\left(0\right)=x-\frac{1}{x^2}\\ -\frac{1}{x^2}+\left(x\right)=x-\frac{1}{x^2}\\ x-\frac{1}{x^2}=x-\frac{1}{x^2}\end{array}$

So, $y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_1$ is the solution of differential equation when $x<0$ .

Case 2: when $x>0$ then

Solution is $y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_2$ .

If $y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_2$ is the solution of differential equation then it will satisfy the differential equation.

  $\begin{array}{c}\frac{dy}{dx}=x-\frac{1}{x^2}\\ \frac{d}{dx}\left(y\right)=x-\frac{1}{x^2}\\ \frac{d}{dx}\left(\frac{1}{x}+\frac{x^2}{2}+C_2\right)=x-\frac{1}{x^2}\left(\text{since }y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_2\right)\\ \frac{d}{dx}\left(\frac{1}{x}\right)+\frac{d}{dx}\left(\frac{x^2}{2}\right)+\frac{d}{dx}\left(C_2\right)=x-\frac{1}{x^2}\\ \frac{d}{dx}\left(x^{-1}\right)+\frac{d}{dx}\left(\frac{x^2}{2}\right)+\frac{d}{dx}\left(C_2\right)=x-\frac{1}{x^2}\\ -1x^{-2}+\left(\frac{2x}{2}\right)+\left(0\right)=x-\frac{1}{x^2}\\ -\frac{1}{x^2}+\left(x\right)=x-\frac{1}{x^2}\\ x-\frac{1}{x^2}=x-\frac{1}{x^2}\end{array}$

So, $y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_2$ is the solution of differential equation when $x>0$ .

Conclusion:

Both the equations $y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_1$ and $y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_2$ are the solutions of differential equation when $x<0$ and $x>0$ respectively.

d.

To determine

To find values for $C_1$ and $C_2$ so that the solution in part (c) agrees with the solutions in parts (a) and (b).

Piecewise function: $y=\left\{\begin{array}{l}\frac{1}{x}+\frac{x^2}{2}+C_{1}x<0\\ \frac{1}{x}+\frac{x^2}{2}+C_{2}x>0\end{array}\right.$

Answer to Problem 75E

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

Explanation of Solution

Given:

Differential equation: $\frac{dy}{dx}=x-\frac{1}{x^2}$

From part (a): Solution is $y(x)=\frac{1}{x}+\frac{x^2}{2}+\frac{1}{2}$ in the interval $\left(0,\infty \right)$ .

From part (b): Solution is $y(x)=\frac{1}{x}+\frac{x^2}{2}+\frac{3}{2}$ in the interval $\left(-\infty ,0\right)$ .

From part (c): Solution of differential equation is piecewise function: $y=\left\{\begin{array}{l}\frac{1}{x}+\frac{x^2}{2}+C_{1}x<0\\ \frac{1}{x}+\frac{x^2}{2}+C_{2}x>0\end{array}\right.$

Calculation:

From part (a) and part (b), the solution of differential equation over the interval $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ is a piecewise function because

From part (a): Solution is $y(x)=\frac{1}{x}+\frac{x^2}{2}+\frac{1}{2}$ in the interval $\left(0,\infty \right)$ .

From part (b): Solution is $y(x)=\frac{1}{x}+\frac{x^2}{2}+\frac{3}{2}$ in the interval $\left(-\infty ,0\right)$ .

So, piecewise function is

  $y=\left\{\begin{array}{l}\frac{1}{x}+\frac{x^2}{2}+\frac{3}{2}x<0\\ \frac{1}{x}+\frac{x^2}{2}+\frac{1}{2}x>0\end{array}\right.$

Conclusion:

Both the equations $y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_1$ and $y\left(x\right)=\frac{1}{x}+\frac{x^2}{2}+C_2$ are the solutions of differential equation when $x<0$ and $x>0$ respectively.