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Q14E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 374
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

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

8s25s+9(s+1)(s23s+2)

The partial fraction expansions for the given rational function 8s25s+9(s+1)(s23s+2) is 1s+111s2+2s1.

See the step by step solution

Step by Step Solution

Step 1: Definition of partial fraction expansion

Any number which can be easily represented in the form of p/q, such that localid="1662726245663" p and localid="1662726249923" q are integers and localid="1662726253823" role="math" q0 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 localid="1662726263552" P and localid="1662726260804" Q are polynomials in localid="1662726257626" x and localid="1662726267995" Q(x)0.

A rational function is known as proper if the degree of localid="1662726271465" 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 P(x)/Q(x) is improper, then it can be expressed as:

P(x)Q(x)=A(x)+R(x)Q(x)

Here,localid="1662726284688" A(x) is a polynomial in localid="1662726275840" x and localid="1662726279992" R(x)/Q(x) is a proper rational function.

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

The given rational function is 8s25s+9(s+1)(s23s+2)

Rewrite 8s25s+9(s+1)(s23s+2) as a sum of partial fractions as:

8s25s+9(s+1)(s2)(s1)=As+1+Bs2+Cs1

Multiply both sides by the LCD (s+1)(s2)(s1) as follows:

8s25s+9=A(s2)(s1)+B(s+1)(s1)+C(s+1)(s2)

Find the constants as:

For s=1,8(1)25(1)+9=A(3)(2)A=1

For s=2:8(2)25(2)+9=B(3)(1)B=11.

For s=1:8(1)25(1)+9=C(2)(1)C=2 .

Substitute the value of constants into 8s25s+9(s+1)(s2)(s1)=As+1+Bs2+Cs1 as follows:

8s25s+9(s+1)(s2)(s1)=1s+1+11s2+2s1=1s+111s2+2s1

Therefore, the partial fraction expansion for the given rational function is 1s+111s2+2s1.

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