Q17E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

\(\frac{{3s + 5}}{{s\left( {{s^2} + s - 6} \right)}}\)

Step-by-Step Solution

Answer

The partial fraction expansions for the given rational function is \(\frac{{3s + 5}}{{s(s + 3)(s - 2)}} =  - \frac{5}{{6s}} - \frac{4}{{15(s + 3)}} + \frac{{11}}{{10(s - 2)}}\)

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 \({\rm{P}}({\rm{x}})/{\rm{Q}}({\rm{x}})\) 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 \(R(x)/Q(x)\) 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

\(\frac{{3s + 5}}{{s\left( {{s^2} + s - 6} \right)}}\)

Factor the denominator

\(\frac{{3s + 5}}{{s\left( {{s^2} + s - 6} \right)}} = \frac{{3s + 5}}{{s(s + 3)(s - 2)}}\)

Write as sum of partial fractions

\(\frac{{3s + 5}}{{s(s + 3)(s - 2)}} = \frac{A}{s} + \frac{B}{{s + 3}} + \frac{C}{{s - 2}}\)

Multiply both sides by the LCD \(s(s + 3)(s - 2)\).

\(3s + 5 = A(s + 3)(s - 2) + Bs(s - 2) + Cs(s + 3)\)

Step 3: Find the constants

Let \(s = 0:3(0) + 5 = A(3)( - 2) \Rightarrow A =  - \frac{5}{6}\)

Let \(s =  - 3:3( - 3) + 5 = B( - 3)( - 5) \Rightarrow B =  - \frac{4}{{15}}\)

Let \(s = 2:3(2) + 5 = C(2)(5) \Rightarrow C = \frac{{11}}{{10}}\)

Substitute the constants:

\(\frac{{3s + 5}}{{s(s + 3)(s - 2)}} =  - \frac{5}{{6s}} - \frac{4}{{15(s + 3)}} + \frac{{11}}{{10(s - 2)}}\)

The partial fraction expansions for the given rational function is \(\frac{{3s + 5}}{{s(s + 3)(s - 2)}} =  - \frac{5}{{6s}} - \frac{4}{{15(s + 3)}} + \frac{{11}}{{10(s - 2)}}\)