Q13E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 11–20, determine the partial fraction expansion for the given rational function.$\frac{\mathbf{-}\mathbf{2}{\mathbf{s}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{s}\mathbf{-}\mathbf{2}}{\mathbf{s}{\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}$

Step-by-Step Solution

Answer

The partial fraction expansion for the given ration function is  $\frac{1}{{(\mathrm{s}+1)}^2}-\frac{2}{\mathrm{s}}$.

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  $\frac{-2{\mathrm{s}}^2-3\mathrm{s}-2}{\mathrm{s}{(\mathrm{s}+1)}^2}$.

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

$\frac{-2{\mathrm{s}}^2-3\mathrm{s}-2}{\mathrm{s}{(\mathrm{s}+1)}^2}=\frac{\mathrm{A}}{\mathrm{s}}+\frac{\mathrm{B}}{(\mathrm{s}+1)}+\frac{\mathrm{C}}{{(\mathrm{s}+1)}^2}$

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

$\begin{array}{c}-2{\mathrm{s}}^2-3\mathrm{s}-2=\mathrm{A}{\left(\mathrm{s}+1\right)}^2+\mathrm{Bs}(\mathrm{s}+1)+\mathrm{Cs}\\ -2{\mathrm{s}}^2-3\mathrm{s}-2={\mathrm{As}}^2+2\mathrm{As}+\mathrm{A}+{\mathrm{Bs}}^2+\mathrm{Bs}+\mathrm{Cs}\end{array}$

Group like terms and factor as:

$\begin{array}{c}-2{\mathrm{s}}^2-3\mathrm{s}-2=({\mathrm{As}}^2+{\mathrm{Bs}}^2)+(2\mathrm{As}+\mathrm{Bs}+\mathrm{Cs})+\mathrm{A}\\ -2{\mathrm{s}}^2-3\mathrm{s}-2=(\mathrm{A}+\mathrm{B}){\mathrm{s}}^2+(2\mathrm{A}+\mathrm{B}+\mathrm{C})\mathrm{s}+\mathrm{A}\end{array}$

Compare coefficients of  ${\mathrm{s}}^2$, $\mathrm{s}$  and constant term as follows:

$\begin{array}{c}\mathrm{A}+\mathrm{B}=-2\\ 2\mathrm{A}+\mathrm{B}+\mathrm{C}=-3\\ \mathrm{A}=-2\end{array}$

 Find the constants as:

$\mathrm{A}=-2,\mathrm{B}=0,\mathrm{C}=1$

Substitute the value of constants into $\frac{-2{\mathrm{s}}^2-3\mathrm{s}-2}{\mathrm{s}{(\mathrm{s}+1)}^2}=\frac{\mathrm{A}}{\mathrm{s}}+\frac{\mathrm{B}}{(\mathrm{s}+1)}+\frac{\mathrm{C}}{{(\mathrm{s}+1)}^2}$  as:

$\begin{array}{l}\frac{-2{\mathrm{s}}^2-3\mathrm{s}-2}{\mathrm{s}{(\mathrm{s}+1)}^2}=\frac{-2}{\mathrm{s}}+\frac{0}{(\mathrm{s}+1)}+\frac{1}{{(\mathrm{s}+1)}^2}\\ \frac{-2{\mathrm{s}}^2-3\mathrm{s}-2}{\mathrm{s}{(\mathrm{s}+1)}^2}=\frac{1}{{(\mathrm{s}+1)}^2}-\frac{2}{\mathrm{s}}\end{array}$

Therefore, the partial fraction expansion for the given ration function is  $\frac{1}{{(\mathrm{s}+1)}^2}-\frac{2}{\mathrm{s}}$.