/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} 34E Theorem 6 in Section 7.3 on page... [FREE SOLUTION] | 91Ó°ÊÓ

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

L-1dnFdsnt=-tnft,

Wheref=L-1F.Use this equation in Problems 33-36 to computeL-1F.Fs=lns-4s-3.

Short Answer

Expert verified

L-1F=1te3t-e4t

Step by step solution

01

Simplify the function and find derivative

Using the property of function lnt we get:

Fs=lns-4s-3Fs=lns-4-lns-3

Find the derivative of F with respect to s:

dFds=ddslns-4-lns-3dFds=ddslns-4-ddslns-3dFds=1s-4-1s-3

02

Find the Laplace inverse

From the given condition, we haveL-1F=1-t1L-1dFdsL-1F=1-tL-11s-4-1s-3L-1F=1t-L-11s-4+1s-3L-1F=1te3t-e4tTherefore,L-1F=1te3t-e4t

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