anonymous
  • anonymous
Laplace Transformation
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[L[e^{2t}tcost]\]
Jack1
  • Jack1
so your initial function is : (e^2t) x (t cos (t))... yeah?
anonymous
  • anonymous
yeah

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Jack1
  • Jack1
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Jack1
  • Jack1
k have a look at this then its from a table of laplace transforms
anonymous
  • anonymous
How to solve it without that?
Jack1
  • Jack1
page 3 and 4 give a pretty good explanation
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Jack1
  • Jack1
for most exam type situations, you will be allowed to take in the table of standard transformations so as to attack the problem in a timely manner if you wish to learn more tho I'd recommend: http://tutorial.math.lamar.edu/Classes/DE/LaplaceIntro.aspx and http://www.sosmath.com/diffeq/laplace/basic/basic.html
John_ES
  • John_ES
You can try to solve the integral, \[F[s]=\int_0^\infty e^{(2-s)t}t\cos t dt\] But, it should be better to solve first, \[f[\alpha]=\int e^{\alpha t}\cos t dt\] Because, \[\int e^{\alpha t}t\cos t dt=\frac{\partial f}{\partial \alpha}\] Following this philoshopy, first, \[f[\alpha]=\int e^{\alpha t}\cos t dt=\frac{e^{\alpha t}}{1+\alpha^2}(\sin t+\alpha \cos t)\] Then \[\frac{\partial f}{\partial \alpha}=\frac{e^{\alpha t}((1+\alpha^2)(t(\sin t+\alpha\cos t)+\cos t)-2\alpha(\sin t+\alpha \cos t))}{(1+\alpha^2)^2}\] Taking limits (exponentials vanish in infinite), you have, \[F[\alpha]=\frac{\alpha^2-1}{(1+\alpha^2)^2}\]And \[\alpha=2-s\] You have, \[F[s]=\frac{s^2-4s+3}{(s^2-4s+5)^2}\]
Jack1
  • Jack1
yeah I'd go with @John_ES 's method lynncake

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