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Chapter 7: 36E (page 350)

Theorem 6 in Section 7.3 on page 364 can be expressed in terms of the inverse Laplace transform as

L-1dnFdsnt=-tnft

Where f=L-1F.Use this equation in Problems 33-36 to compute

L-1F.role="math" localid="1664422930667" Fs=arctan1s

Short Answer

Expert verified

Fs=arctan1s

Step by step solution

01

Simplify the function and find derivative 

Find the derivative of F with respect to s:

dFds=ddsarctan1s=11+1s2-1s2=-1s2+1

02

Find the Laplace inverse

From the given condition, we have L-1F=1-tnL-1dnFdsn,apply this to find the Laplace inverse as:

L-1F=1-t1L-1dFdsL-1F=1-tL-1-1s2+1=--1tL-11s2+1L-1F=1tsint

Therefore,

L-1F=1tsint

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Most popular questions from this chapter

In Problems 1-20, determine the Laplace transform of the given function using Table 7.1 on page 356 and the properties of the transform given in Table 7.2. [Hint: In Problems 12-20, use an appropriate trigonometric identity.]


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