Chapter 10.7, Problem 45E

To determine

The values of $A$ and $B$ need to be defined in terms of $a$ and $b$ by using the following eqns:

  $\frac{ax+b}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}$

Answer to Problem 45E

The values of $A$ and $B$ in terms of $a$ and $b$ are $\left(A,B\right)=\left(\frac{a+b}{2},\frac{a-b}{2}\right)$

Explanation of Solution

Given:

  $\frac{ax+b}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}$

Calculations:

The given eq. is:

  $\begin{array}{l}\frac{ax+b}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}\\ \frac{ax+b}{x^2-1}=\frac{A\left(x+1\right)+B\left(x-1\right)}{x^2-1}\end{array}$

From above eq.:

The denominators are equal, so need to use the equality for the numerators as well:

  $\begin{array}{l}ax+b=A\left(x+1\right)+B\left(x-1\right)\\ ax+b=Ax+A+Bx-B\\ ax+b=x\left(A+B\right)+A-B\end{array}$

The coefficients near the like terms should be equal, so the following system is obtained:

  $\begin{array}{l}A+B=a\mathrm{...}\left(1\right)\\ A-B=b\mathrm{...}\left(2\right)\end{array}$

Addition of above two eqns.:

  $\begin{array}{l}A+B+A-B=a+b\\ 2A=a+b\\ A=\frac{a+b}{2}\end{array}$

Now, to find the value of $B$ use eq. (1):

  $\begin{array}{l}A+B=a\\ \frac{a+b}{2}+B=a\\ B=a-\frac{a+b}{2}\\ B=\frac{2a-a-b}{2}\\ B=\frac{a-b}{2}\end{array}$

Conclusion:

Hence, the solution of the given eq. is $\left(A,B\right)=\left(\frac{a+b}{2},\frac{a-b}{2}\right)$ .