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anonymous
 one year ago
guys the inverse laplace transform of e^(sc+k)x is what???
anonymous
 one year ago
guys the inverse laplace transform of e^(sc+k)x is what???

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's the question? To explain it's meaning?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2\[\mathcal L^{1}\big\{e^{(sc+k)x}\big\} \]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2use the shift theorem(s)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then how we use shift thm?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u apply the shift thm on this question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Have you tried it already?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is right or wrong? if right then plz find the inverse laplace of the last term.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0some one check this qstn plz.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0just 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
 one year ago
Best ResponseYou've already chosen the best response.0i am notgetting u.what r u saying??

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0this 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
 one year ago
Best ResponseYou've already chosen the best response.0sin(tcx) this term correct r u sure?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0test 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%28kx%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 retest 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
 one year ago
Best ResponseYou've already chosen the best response.0how we can take inverse laplace of the last statement?
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