$$\huge \mathcal L \lbrace t^2 \cos kt \rbrace$$

i got as far as $$\mathcal L \lbrace t^2 \cos kt \rbrace = \frac{(s^2+k^2)^2 (-2s) - 4s(k^2 - s^2)(k^2 +s^2)}{(s^2 + k^2)^4}$$
what do i do next?

Question

$$\huge \mathcal L \lbrace t^2 \cos kt \rbrace$$

i got as far as $$\mathcal L \lbrace t^2 \cos kt \rbrace = \frac{(s^2+k^2)^2 (-2s) - 4s(k^2 - s^2)(k^2 +s^2)}{(s^2 + k^2)^4}$$
what do i do next?

unklerhaukus · Answer

cancel common factors?

lgbasallote · Answer

i got $$\Large\frac{-2s(3k^2 -s^2)}{(s^2 + k^2)^3}$$

anonymous · Answer

yeah

lgbasallote · Answer

that's right? sweet

lgbasallote · Answer

so wait...that's right? no joke?

anonymous · Answer

it is ok .i think in the out side it should be -2s^2 not 2s . otherwise it looks fine.

lgbasallote · Answer

why :O

lgbasallote · Answer

are you sure it's squared o.O

anonymous · Answer

but with what igba wrote in his post after simplifynig i got -2s(3k^2-s^2)

anonymous · Answer

yes i am sure. what your book says ?

lgbasallote · Answer

none. this is one of those with no answers

lgbasallote · Answer

$$\mathcal L \lbrace \cos kt \rbrace = \frac{s}{s^2 + k^2}$$
right?

lgbasallote · Answer

if yes then it seems wolfram agrees with me :DDD http://www.wolframalpha.com/input/?i=second+derivative+of+%28s%2F%28s%5E2+%2B+k%5E2%29%29

anonymous · Answer

yes it is the cos transformation

after first derivative you should have
$$\Large \frac{k^2-s^2}{(s^2+k^2)^2}$$
right ?

anonymous · Answer

lol http://www.wolframalpha.com/input/?i=+laplace+transform+of+%28t%5E2+cos%28+kt%29%29

lgbasallote · Answer

weird...wolfram agrees with my second derivative...but it agrees with your laplace...

anonymous · Answer

sorry guys i did it :P you were right :P

lgbasallote · Answer

lol why did you do $$\large \mathcal L \lbrace t^2 s \cos kt \rbrace$$
that's cheating :p

anonymous · Answer

i did it on paper as well :P
look above i gave you first derivative :)