Question

guys the inverse laplace transform of e^-(sc+k)x is what???

Answer 1

What's the question? To explain it's meaning?

Answer 2

\[\mathcal L^{-1}\big\{e^{-(sc+k)x}\big\} \]

Answer 3

use the shift theorem(s)

Answer 4

yes

Answer 5

then how we use shift thm?

Answer 6

can u apply the shift thm on this question?

Answer 7

Have you tried it already?

Answer 8

this is right or wrong? if right then plz find the inverse laplace of the last term.

Answer 9

some one check this qstn plz.

Answer 10

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.

Answer 11

thank u.

Answer 12

i am notgetting u.what r u saying??

Answer 13

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.

Answer 14

sin(t-cx) this term correct r u sure?

Answer 15

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.

Answer 16

how we can take inverse laplace of the last statement?