39E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the residue computation formula derived in Problem 38 to determine quickly the partial fraction expansion for$F\left(s\right)=\frac{2s+1}{s\left(s-1\right)\left(s+2\right)}$

Step-by-Step Solution

Answer

$\frac{2s+1}{s\left(s-1\right)\left(s+2\right)}=-\frac{1}{2s}+\frac{1}{s-1}-\frac{1}{2\left(s+2\right)}$

Step 1: Define Inverse Laplace transform

Given a function  , if there is a function  that is continuous on

 and satisfies,then we say that  is the inverse Laplace transform of and employ the notation

Non-repeated Linear Factors

If  can be factored into a product of distinct linear factors,

where the   's are all distinct real numbers, then the partial fraction expansion has the form

where the   's are real numbers. There are various ways of determining the constants . In the next example, we demonstrate two such methods.2.

Repeated Linear Factors

If   is a factor of   and   is the highest power of   that divides , then the portion of the partial fraction expansion of   that corresponds to the term   is

where the 's are real numbers.

Quadratic Factors

If   is a quadratic factor of   that cannot be reduced to linear factors with real coefficients and    is the highest power of   that divides  , then the portion of the partial fraction expansion that corresponds to    is

it is more convenient to express  in the form when we look up the Laplace transforms. So let's agree to write this portion of the partial fraction expansion in the equivalent form

Step 2: Find the factor of the denominator

There we have

$P\left(s\right)=2s+1,Q\left(s\right)=s\left(s-1\right)\left(s+2\right)$

Using partial fractions we get

$F\left(s\right)=\frac{2s+1}{s\left(s-1\right)\left(s+2\right)}=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{2\left(s+2\right)}$

and using the equation from Problem 28 we obtain

$$\begin{gathered} A=\underset{s\to 0}{\lim }sF\left(s\right) \\ =\frac{2s+1}{\left(s-1\right)\left(s+2\right)} \\ =-\frac{1}{2} \end{gathered}$$

$$\begin{gathered} B=\underset{s\to 1}{\lim }sF\left(s\right) \\ =\underset{s\to 1}{\lim }\frac{2s+1}{s\left(s+2\right)} \\ =1 \end{gathered}$$

$$\begin{gathered} C=\underset{s\to -2}{\lim }sF\left(s\right) \\ =\underset{s\to -2}{\lim }\frac{2s+1}{s\left(s-2\right)} \\ =-\frac{1}{2} \end{gathered}$$

Therefore

$\frac{2s+1}{s\left(s-1\right)\left(s+2\right)}=-\frac{1}{2s}+\frac{1}{s-1}-\frac{1}{2\left(s+2\right)}$