Q11E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 11–20, determine the partial fraction expansion for the given rational function.$\frac{{\mathbf{s}}^{\mathbf{2}}\mathbf{-}\mathbf{26}\mathbf{s}\mathbf{-}\mathbf{47}}{\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{5}\mathbf{)}}$

Step-by-Step Solution

Answer

The partial fraction expansion for the given rational function is $\frac{6}{\mathrm{s}+5}-\frac{1}{\mathrm{s}+2}-\frac{4}{\mathrm{s}-1}$.

Step 1: Definition of partial fraction expansion

• The partial fraction expansion of a rational fraction (that is, a fraction with both a polynomial numerator and a polynomial denominator) is an algebraic procedure that involves expressing the fraction as a sum of a polynomial plus one or more fractions with a simpler denominator.
• The partial fraction decomposition is important because it gives methods for a variety of rational function computations, such as explicit antiderivative computations, Taylor series expansions, inverse Z-transforms, and inverse Laplace transforms.

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

The given rational function is  $\frac{{\mathrm{s}}^2-26\mathrm{s}-47}{(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)}$

Rewrite  $\frac{{\mathrm{s}}^2-26\mathrm{s}-47}{(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)}$ as a sum of partial fractions as:

$\frac{{\mathrm{s}}^2-26\mathrm{s}-47}{(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)}=\frac{\mathrm{A}}{\mathrm{s}-1}+\frac{\mathrm{B}}{\mathrm{s}+2}+\frac{\mathrm{C}}{\mathrm{s}+5}$

Multiply both sides by the LCD $(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)$  as follows:

 ${\mathrm{s}}^2-26\mathrm{s}-47=\mathrm{A}(\mathrm{s}+2)(\mathrm{s}+5)+\mathrm{B}(\mathrm{s}-1)(\mathrm{s}+5)+\mathrm{C}(\mathrm{s}-1)(\mathrm{s}+2)$

Find the constants as:

For  $\mathrm{s}=1,{\left(1\right)}^2-26\left(1\right)-47=\mathrm{A}\left(3\right)\left(6\right)\Rightarrow \mathrm{A}=-4$.

 

For  $\mathrm{s}=-2,{(-2)}^2-26(-2)-47=\mathrm{B}(-3)\left(3\right)\Rightarrow \mathrm{B}=-1$.

 

For  $\mathrm{s}=-5,{(-5)}^2-26(-5)-47=\mathrm{C}(-6)(-3)\Rightarrow \mathrm{C}=6$ .

Substitute the value of constants into $\frac{{\mathrm{s}}^2-26\mathrm{s}-47}{(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)}=\frac{\mathrm{A}}{\mathrm{s}-1}+\frac{\mathrm{B}}{\mathrm{s}+2}+\frac{\mathrm{C}}{\mathrm{s}+5}$  as follows:

$\begin{array}{c}\frac{{\mathrm{s}}^2-26\mathrm{s}-47}{(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)}=\frac{-4}{\mathrm{s}-1}+\frac{-1}{\mathrm{s}+2}+\frac{6}{\mathrm{s}+5}\\ =\frac{6}{\mathrm{s}+5}-\frac{1}{\mathrm{s}+2}-\frac{4}{\mathrm{s}-1}\end{array}$

Therefore, the partial fraction expansion for the given rational function is 

$\frac{6}{\mathrm{s}+5}-\frac{1}{\mathrm{s}+2}-\frac{4}{\mathrm{s}-1}$