Chapter 7.5, Problem 17E

To determine

To solve the differential equation

Answer to Problem 17E

The general solution of differential equation is $F\left(x\right)=-2\ln \left|x\right|+\ln \left|x+1\right|+\ln \left|x-1\right|+C$ .

Explanation of Solution

Given information:

The differential equation is $F^{\prime }\left(x\right)=\frac{2}{x^3-x}$

Calculation:

The differential equation is $F^{\prime }\left(x\right)=\frac{2}{x^3-x}$

Using variable separable method to solve differential equation

  $\begin{array}{l}{F}^{\prime }\left(x\right)=\frac{2}{x^3-x}\\ \frac{dy}{dx}=\frac{2}{x^3-x}\\ dy=\left(\frac{2}{x^3-x}\right)dx\\ \text{integrating both side}\end{array}$

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

  $\frac{2}{x^3-x}=\frac{2}{x\left(x+1\right)\left(x-1\right)}\text{=}\frac{A}{x}\text{+}\frac{B}{x+1}\text{+}\frac{C}{x-1}\text{=}\frac{A\left(x^2-1\right)+B\left(x^2-x\right)+C\left(x^2+x\right)}{x\left(x+1\right)\left(x-1\right)}$

  $2=A\left(x^2-1\right)+B\left(x^2-x\right)+C\left(x^2+x\right)=x^2\left(A+B+C\right)+x\left(-B+C\right)-A$

  $\begin{array}{l}\text{Compairing constant term}\\ -A=2\\ A=-2\\ \text{Compairing coefficent of }x\\ -B+C=0\\ \text{Equation 1 : }-B+C=0\\ \text{Compairing coefficent of }{x}^2\end{array}$

  $\begin{array}{l}A+B+C=0\\ B+C=-A\\ B+C=2\\ \text{Equation 2 : }B+C=2\end{array}$

  $\begin{array}{l}\text{Now , adding equation 1 and equation 2}\\ \text{2}C=2\\ C=1\\ -B+C=0\\ B=C=1\end{array}$

  $\therefore y={\displaystyle \int \left(\frac{2}{x^3-x}\right)dx}={\displaystyle \int \left(\frac{-2}{x}\text{+}\frac{1}{x+1}\text{+}\frac{1}{x-1}\right)dx}=-2\ln \left|x\right|+\ln \left|x+1\right|+\ln \left|x-1\right|+C\text{, where }C\text{ is constant}$

  $F\left(x\right)=-2\ln \left|x\right|+\ln \left|x+1\right|+\ln \left|x-1\right|+C$

The general solution of differential equation is $F\left(x\right)=-2\ln \left|x\right|+\ln \left|x+1\right|+\ln \left|x-1\right|+C$ .