Solution Found!
Solved: Theorem 6 in Section 7.3 can be expressed in terms
Chapter 7, Problem 33E(choose chapter or problem)
Theorem 6 in Section 7.3 can be expressed in terms of the inverse Laplace transform as
\(\mathscr{\mathcal { L }}^{1}\left\{\frac{d^{n} F}{d s^{n}}\right\}(t)=(-t)^{n} f(t)\)
where \(\mathscr{\mathscr { L }}^{-1}\{F)\). Use this equation in Problems 33–36 to compute \(\mathscr{L}^{-1}\{F\}\).
\(F(s)=\ln \left(\frac{s+2}{s-5}\right)\)
Equation transcription:
Text transcription:
{ { L }}^{1}{frac{d^{n} F}{d s^{n}}}(t)=(-t)^{n} f(t)
f=r{L}^{-1}{F}
{{ L }}^{-1}\{F)
{L}^{-1}{F\}
F(s)=ln (\frac{s+2}{s-5})
Questions & Answers
QUESTION:
Theorem 6 in Section 7.3 can be expressed in terms of the inverse Laplace transform as
\(\mathscr{\mathcal { L }}^{1}\left\{\frac{d^{n} F}{d s^{n}}\right\}(t)=(-t)^{n} f(t)\)
where \(\mathscr{\mathscr { L }}^{-1}\{F)\). Use this equation in Problems 33–36 to compute \(\mathscr{L}^{-1}\{F\}\).
\(F(s)=\ln \left(\frac{s+2}{s-5}\right)\)
Equation transcription:
Text transcription:
{ { L }}^{1}{frac{d^{n} F}{d s^{n}}}(t)=(-t)^{n} f(t)
f=r{L}^{-1}{F}
{{ L }}^{-1}\{F)
{L}^{-1}{F\}
F(s)=ln (\frac{s+2}{s-5})
ANSWER:Solution:
Step 1:
In this problem we need to determine .