Q18E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

\(\frac{{3{s^2} + 5s + 3}}{{{s^4} + {s^3}}}\)

Step-by-Step Solution

Answer

The partial fraction expansions for the given rational function is \(\frac{{3{s^2} + 5s + 3}}{{{s^3}\left( {s + 1} \right)}} = \frac{1}{s} + \frac{2}{{{s^2}}} + \frac{3}{{{s^3}}} - \frac{1}{{s + 1}}\)

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

\(\frac{{3{s^2} + 5s + 3}}{{{s^4} + {s^3}}} = \frac{{3{s^2} + 5s + 3}}{{{s^3}\left( {s + 1} \right)}}\)

Write as sum of partial fractions

\(\frac{{3{s^2} + 5s + 3}}{{{s^3}\left( {s + 1} \right)}} = \frac{A}{s} + \frac{B}{{{s^2}}} + \frac{C}{{{s^3}}} + \frac{D}{{s + 1}}\)

Multiply both sides by the \({\rm{LCD}}\;{s^3}\left( {s + 1} \right)\)

\(3{s^2} + 5s + 3 = A{s^2}\left( {s + 1} \right) + Bs\left( {s + 1} \right) + C\left( {s + 1} \right) + D{s^3}\)

Simplify:

\(3{s^2} + 5s + 3 = A{s^3} + A{s^2} + B{s^2} + Bs + Cs + C + D{s^3}\)

Group like terms and factor:

\(\begin{array}{l}3{s^2} + 5s + 3 = \left( {A{s^3} + D{s^3}} \right) + \left( {A{s^2} + B{s^2}} \right) + \left( {Bs + Cs} \right) + C\\3{s^2} + 5s + 3 = \left( {A + D} \right){s^3} + \left( {A + B} \right){s^2} + \left( {B + C} \right)s + C\end{array}\)

Compare coefficients:

\(3{s^2} + 5s + 3 = \underbrace {\left( {A + D} \right)}_0{s^3} + \underbrace {\left( {A + B} \right)}_3{s^2} + \underbrace {\left( {B + C} \right)}_5s + \underbrace C_3\)

 

Step 3: Find the constants

Therefore,

\(\left\{ {\begin{array}{*{20}{c}}{A + D = 0}\\{A + B = 3}\\{B + C = 5}\\{\;\;\;\;\;C = 3}\end{array}} \right.\)

Find the constants, solve the system, you get:

\(A = 1,B = 2,C = 3,D =  - 1\)

Substitute the constants:

\(\begin{array}{l}\frac{{3{s^2} + 5s + 3}}{{{s^3}\left( {s + 1} \right)}} = \frac{A}{s} + \frac{B}{{{s^2}}} + \frac{C}{{{s^3}}} + \frac{D}{{s + 1}}\\\frac{{3{s^2} + 5s + 3}}{{{s^3}\left( {s + 1} \right)}} = \frac{1}{s} + \frac{2}{{{s^2}}} + \frac{3}{{{s^3}}} - \frac{1}{{s + 1}}\end{array}\)

The partial fraction expansions for the given rational function is \(\frac{{3{s^2} + 5s + 3}}{{{s^3}\left( {s + 1} \right)}} = \frac{1}{s} + \frac{2}{{{s^2}}} + \frac{3}{{{s^3}}} - \frac{1}{{s + 1}}\).