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

1. anonymous

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

2. UnkleRhaukus

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

3. UnkleRhaukus

use the shift theorem(s)

4. anonymous

yes

5. anonymous

then how we use shift thm?

6. anonymous

can u apply the shift thm on this question?

7. anonymous

8. anonymous

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

9. anonymous

some one check this qstn plz.

10. 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.

11. anonymous

thank u.

12. anonymous

i am notgetting u.what r u saying??

13. 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.

14. anonymous

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

15. 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.

16. anonymous

how we can take inverse laplace of the last statement?