Q14E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\frac{\mathbf{-}\mathbf{8}{\mathbf{s}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathbf{s}\mathbf{+}\mathbf{9}}{\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}{\mathbf{s}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{s}\mathbf{+}\mathbf{2}\mathbf{)}}$

Step-by-Step Solution

Answer

The partial fraction expansions for the given rational function $\frac{-8{\mathrm{s}}^2-5\mathrm{s}+9}{(\mathrm{s}+1)({\mathrm{s}}^2-3\mathrm{s}+2)}$  is $\frac{1}{\mathrm{s}+1}-\frac{11}{\mathrm{s}-2}+\frac{2}{\mathrm{s}-1}$.

Step 1: Definition of partial fraction expansion

Any number which can be easily represented in the form of $\mathrm{p}/\mathrm{q}$, such that $\mathrm{p}$ and $\mathrm{q}$ are integers and $\mathrm{q}\ne 0$  is known as a rational number.

Similarly, we can define a rational function as the ratio of two polynomial functions $\mathrm{P}\left(\mathrm{x}\right)$ and $\mathrm{Q}\left(\mathrm{x}\right)$ , where $\mathrm{P}$ and $\mathrm{Q}$ are polynomials in $\mathrm{x}$  and  $\mathrm{Q}\left(\mathrm{x}\right)\ne 0$.

A rational function is known as proper if the degree of  $\mathrm{P}\left(\mathrm{x}\right)$ is less than the degree of  $\mathrm{Q}\left(\mathrm{x}\right)$; 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  $\text{P}\left(\text{x}\right)/\text{Q}\left(\text{x}\right)$ is improper, then it can be expressed as:

$\frac{\mathrm{P}\left(\mathrm{x}\right)}{\mathrm{Q}\left(\mathrm{x}\right)}=\mathrm{A}\left(\mathrm{x}\right)+\frac{\mathrm{R}\left(\mathrm{x}\right)}{\mathrm{Q}\left(\mathrm{x}\right)}$

Here,$\mathrm{A}\left(\mathrm{x}\right)$ is a polynomial in $\mathrm{x}$ and $\mathrm{R}\left(\mathrm{x}\right)/\mathrm{Q}\left(\mathrm{x}\right)$ is a proper rational function.

Step 2: Determine the partial fraction expansion for the given rational function

The given rational function is $\frac{-8{\mathrm{s}}^2-5\mathrm{s}+9}{(\mathrm{s}+1)({\mathrm{s}}^2-3\mathrm{s}+2)}$

Rewrite $\frac{-8{\mathrm{s}}^2-5\mathrm{s}+9}{(\mathrm{s}+1)({\mathrm{s}}^2-3\mathrm{s}+2)}$ as a sum of partial fractions as:

$\frac{-8{\mathrm{s}}^2-5\mathrm{s}+9}{(\mathrm{s}+1)(\mathrm{s}-2)(\mathrm{s}-1)}=\frac{\mathrm{A}}{\mathrm{s}+1}+\frac{\mathrm{B}}{\mathrm{s}-2}+\frac{\mathrm{C}}{\mathrm{s}-1}$

Multiply both sides by the LCD $(\mathrm{s}+1)(\mathrm{s}-2)(\mathrm{s}-1)$  as follows:

$-8{\mathrm{s}}^2-5\mathrm{s}+9=\mathrm{A}(\mathrm{s}-2)(\mathrm{s}-1)+\mathrm{B}(\mathrm{s}+1)(\mathrm{s}-1)+\mathrm{C}(\mathrm{s}+1)(\mathrm{s}-2)$

Find the constants as:

For $\mathrm{s}=-1,-8{(-1)}^2-5(-1)+9=\mathrm{A}(-3)(-2)\Rightarrow \mathrm{A}=1$

For $\mathrm{s}=2:-8{\left(2\right)}^2-5\left(2\right)+9=\mathrm{B}\left(3\right)\left(1\right)\Rightarrow \mathrm{B}=-11$.

 

For $\mathrm{s}=1:-8{\left(1\right)}^2-5\left(1\right)+9=\mathrm{C}\left(2\right)(-1)\Rightarrow \mathrm{C}=2$ .

Substitute the value of constants into $\frac{-8{\mathrm{s}}^2-5\mathrm{s}+9}{(\mathrm{s}+1)(\mathrm{s}-2)(\mathrm{s}-1)}=\frac{\mathrm{A}}{\mathrm{s}+1}+\frac{\mathrm{B}}{\mathrm{s}-2}+\frac{\mathrm{C}}{\mathrm{s}-1}$  as follows:

$\begin{array}{c}\frac{-8{\mathrm{s}}^2-5\mathrm{s}+9}{(\mathrm{s}+1)(\mathrm{s}-2)(\mathrm{s}-1)}=\frac{1}{\mathrm{s}+1}+\frac{-11}{\mathrm{s}-2}+\frac{2}{\mathrm{s}-1}\\ =\frac{1}{\mathrm{s}+1}-\frac{11}{\mathrm{s}-2}+\frac{2}{\mathrm{s}-1}\end{array}$

Therefore, the partial fraction expansion for the given rational function is  $\frac{1}{\mathrm{s}+1}-\frac{11}{\mathrm{s}-2}+\frac{2}{\mathrm{s}-1}$.