Question

Laplace transform for f(t)=tcosh(3t)?

Answer 1

I looked at the table and it says (-1)^n*F^(n) of (s)

Answer 2

the laplace transform of cosh(3t) is s/(s^2-9) so I thought it was that multiplied by -1

Answer 3

but why is the answer taking the derivative of it

Answer 4

No, because L[fg] does not equal L[f]L[g] in general. (That's the formula for convolution.)

I'd write cosh(3t) = 1/2 (e^3t + e^-3t).

Then find the Laplace transform of te^at and then use linear superposition.

Answer 5

I don't understand.. is 1/2 (e^3t + e^-3t) that just another way to write cosh(3t)?

Answer 6

yes, that's actually the definition of cosh

Answer 7

and sinh(at) = 1/2 (e^at - e^-at)

Answer 8

Oh I guess that works too, but could you tell me why they took the different in this solution?

Answer 9

derivative**

Answer 10

Yes, it's because
\[ L[ t g(t) ] = - G'(s) \]
where \( L[g(t)] = G(s) \)

That identity is true because
\[ \frac{d \ }{ds} G(s) = \frac{d \ }{ds} \int_0^\infty g(t) e^{-st} dt \]
\[ = \int_0^\infty \frac{\partial \ }{\partial s} ( g(t) e^{-st}) \ dt \]
\[ = \int_0^\infty g(t) \frac{\partial \ }{\partial s} e^{-st} \ dt \]
\[ = \int_0^\infty g(t) . (-t) e^{-st} \ dt \]
\[ = -\int_0^\infty t g(t) . e^{-st} \ dt \]
\[ = -L[tg(t)] \]

Answer 11

Oh I see it make sense now thanks! I actually have another question but I'll post it in a new board. Thanks agian