anonymous
  • anonymous
guys the inverse laplace transform of e^-(sc+k)x is what???
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
What's the question? To explain it's meaning?
UnkleRhaukus
  • UnkleRhaukus
\[\mathcal L^{-1}\big\{e^{-(sc+k)x}\big\} \]
UnkleRhaukus
  • UnkleRhaukus
use the shift theorem(s)

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anonymous
  • anonymous
yes
anonymous
  • anonymous
then how we use shift thm?
anonymous
  • anonymous
can u apply the shift thm on this question?
anonymous
  • anonymous
Have you tried it already?
anonymous
  • anonymous
this is right or wrong? if right then plz find the inverse laplace of the last term.
anonymous
  • anonymous
some one check this qstn plz.
IrishBoy123
  • IrishBoy123
just a thought but as you are inverse transforming with respect to \(s\), should you not: \( e^{-(sc+k)x} = e^{-(cx)s}.e^{-kx} \) which makes it look like this Heaviside shift: \( e^{-kx} . H(t - cx) . sin (t - cx) \) rest looks OK to me.
anonymous
  • anonymous
thank u.
anonymous
  • anonymous
i am notgetting u.what r u saying??
IrishBoy123
  • IrishBoy123
this is not a convolution but a Heaviside shift \(e^{-kx}\) is effectively a constant in the inverse transformation, so you are looking at \((e^{-kx}\mathcal{L^{-1}}\{ e^{-(cx)s} \} \), which represents a Heaviside shift of cx. the solution i posted meets the IV's, if you plug them in. you can also plug the solution back into the original DE and you will see that it works.
anonymous
  • anonymous
sin(t-cx) this term correct r u sure?
IrishBoy123
  • IrishBoy123
test it. i couldn't get Wolfram to solve the PDE (maybe you need paid membership for that) but it seems to allow you to work backwards, eg http://www.wolframalpha.com/input/?i=d2%2Fdx%5E2+%5B+e%5E%28-kx%29+*+sin+%28t+-+cx%29%5D so, if i was unconvinced, i'd do this for each term in the original equation and add them up. you should also as a matter of course re-test the IV's as the solution clearly needs to pass that test too. this solution should. then, if and when that is all done and dusted, i'd then wonder why i was doubting the solution.
anonymous
  • anonymous
how we can take inverse laplace of the last statement?

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