Q20E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 11–20, determine the partial fraction expansions for the given rational function.

\(\frac{s}{{(s - 1)\left( {{s^2} - 1} \right)}}\)

Step-by-Step Solution

Answer

The partial fraction expansions for the given rational function is 

\(\frac{s}{{\left( {s + 1} \right){{\left( {s - 1} \right)}^2}}} =   - \frac{1}{{4\left( {s + 1} \right)}} + \frac{1}{{4\left( {s - 1} \right)}} + \frac{1}{{2{{\left( {s - 1} \right)}^2}}}\)

Step 1: Define partial fractions expansions

Any number which can be easily represented in the form of \(p/q\), such that \(p\) and \(q\) are integers and \(q \ne 0\) is known as a rational number. Similarly, we can define a rational function as the ratio of two polynomial functions \(P(x)\) and \(Q(x)\), where \(P\) and \(Q\) are polynomials in \(x\) and \(Q(x) \ne 0\). A rational function is known as proper if the degree of \(P(x)\) is less than the degree of \(Q(x)\); otherwise, it is known as an improper rational function. With the help of the long division process, we can reduce improper rational functions to proper rational functions. Therefore, if \(\frac{{{\rm{P}}({\rm{x}})}}{{{\rm{Q}}({\rm{x}})}}\) is improper, then it can be expressed as:

\(\frac{{P(x)}}{{Q(x)}} = A(x) + \frac{{R(x)}}{{Q(x)}}\)

Here, \(A(x)\) is a polynomial in \(x\) and \(\frac{{R(x)}}{{Q(x)}}\) is a proper rational function.

We know that the integration of a function \(f(x)\) is given by \(F(x)\) and it is represented by:

\(\int f (x)dx = F(x) + C\)

Here R.H.S. of the equation means integral of \(f(x)\) with respect to \(x\) and \(C\) is the constant of integration. 

Step 2: Find the factor of the denominator

Factor the denominator

\(\begin{array}{c}\frac{s}{{(s - 1)\left( {{s^2} - 1} \right)}} = \frac{s}{{(s - 1)(s + 1)(s - 1)}}\\ = \frac{s}{{(s + 1){{(s - 1)}^2}}}\end{array}\)

Write as sum of partial fractions

\(\frac{s}{{\left( {s + 1} \right){{\left( {s - 1} \right)}^2}}} = \frac{A}{{s + 1}} + \frac{B}{{\left( {s - 1} \right)}} + \frac{C}{{{{\left( {s - 1} \right)}^2}}}\)

Multiply both sides by the LCD \(\left( {s + 1} \right){\left( {s - 1} \right)^2}\).

\(\begin{array}{c}s = A{\left( {s - 1} \right)^2} + B\left( {s + 1} \right)\left( {s - 1} \right) + C\left( {s + 1} \right)\\s = A{s^2} - 2As + A + B{s^2} - B + Cs + C\end{array}\)

 

Group like terms and factor. 

\(\begin{array}{l}s = \left( {A{s^2} + B{s^2}} \right) + \left( { - 2As + Cs} \right) + \left( {A - B + C} \right)\\s = \left( {A + B} \right){s^2} + \left( { - 2A + C} \right)s + \left( {A - B + C} \right)\end{array}\)

 

Compare coefficients:  

 

\(s = \underbrace {\left( {A + B} \right)}_0{s^2} + \underbrace {\left( { - 2A + C} \right)}_1s + \underbrace {\left( {A - B + C} \right)}_0\)

 

Step 3: Find the constants

Therefore,

\(\left\{ {\begin{array}{*{20}{c}}{A + B = 0}\\{ - 2A + C = 1}\\{A - B + C = 0}\end{array}} \right.\)

Find the constants

\(A =   - \frac{1}{4},B = \frac{1}{4},C = \frac{1}{2}\)

Substitute the constants:

\(\frac{s}{{\left( {s + 1} \right){{\left( {s - 1} \right)}^2}}} =   - \frac{1}{{4\left( {s + 1} \right)}} + \frac{1}{{4\left( {s - 1} \right)}} + \frac{1}{{2{{\left( {s - 1} \right)}^2}}}\)

The partial fraction expansions for the given rational function is 

\(\frac{s}{{\left( {s + 1} \right){{\left( {s - 1} \right)}^2}}} =   - \frac{1}{{4\left( {s + 1} \right)}} + \frac{1}{{4\left( {s - 1} \right)}} + \frac{1}{{2{{\left( {s - 1} \right)}^2}}}\)