Chapter 6.5, Problem 17E

To determine

To find : The value of the differential equation.

Answer to Problem 17E

The answer is:

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

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}F(x)=\int \frac{2}{x^3-x}dx\\ \frac{2}{x^3-x}=\frac{2}{x\left(x^2-1\right)}\\ =\frac{2}{x(x+1)(x-1)}\\ \frac{2}{x^3-x}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}\\ =\frac{A(x+1)(x-1)}{x(x+1)(x-1)}+\frac{Bx(x-1)}{x(x-1)(x+1)}+\frac{Cx(x+1)}{x(x+1)(x-1)}\\ =\frac{A(x+1)(x-1)+Bx(x-1)+Cx(x+1)}{x(x+1)(x-1)}\\ =\frac{A(x+1)(x-1)+Bx(x-1)+Cx(x+1)}{x^3-x}\end{array}$

So it follows that,

  $A(x+1)(x-1)+Bx(x-1)+Cx(x+1)=\mathrm{2.....}(1)$

Put, $x=1$ in equation $(1)$ and get,

  $\begin{array}{c}A(1+1)(1-1)+B\cdot 1\cdot (1-1)+C\cdot 1\cdot (1+1)=2\\ \Rightarrow A\cdot 0+B\cdot 0+C\cdot 2=2\\ \Rightarrow 2C=2\\ \Rightarrow C=1\end{array}$

Put, $x=-1$ in equation $(1)$ and get,

  $\begin{array}{c}A(-1+1)(-1-1)+B\cdot (-1)\cdot (-1-1)+C\cdot (-1)\cdot (-1+1)=2\\ \Rightarrow A\cdot 0+B\cdot 2+C\cdot 0=2\\ \Rightarrow 2B=2\\ \Rightarrow B=1\end{array}$

Put, $x=0$ in equation $(1)$ and get,

  $\begin{array}{c}A(0+1)(0-1)+B\cdot 0\cdot (0-1)+C\cdot 0\cdot (0+1)=2\\ \Rightarrow A\cdot (-1)+B\cdot 0+C\cdot 0=2\\ \Rightarrow -A=2\\ \Rightarrow A=-2\end{array}$

Therefore, $\frac{2}{x^3-x}=\frac{-2}{x}+\frac{1}{x+1}+\frac{1}{x-1}$ . Thus,

  $\begin{array}{c}F(x)=\int \frac{2}{x^3-x}dx\\ =\int \left(\frac{-2}{x}+\frac{1}{x+1}+\frac{1}{x-1}\right)dx\\ =\int \frac{-2}{x}dx+\int \frac{1}{x+1}dx+\int \frac{1}{x-1}dx\\ =-2\int \frac{1}{x}dx+\int \frac{1}{x+1}dx+\int \frac{1}{x-1}dx\\ =-2\ln \left|x\right|+\ln |x+1|+\ln |x-1|+C\\ =-2\ln \left|x\right|+\ln |(x+1)(x-1)|+C\\ =\ln |x{|}^{-2}+\ln \left|x^2-1\right|+C\\ =\ln \frac{1}{x^2}+\ln \left|x^2-1\right|+C\\ =\ln \frac{\left|x^2-1\right|}{x^2}+C\end{array}$

Therefore the required value of the differential equation $F^{\prime }(x)=\frac{2}{x^3-x}$ is

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