if i try to write out what i did on this it will be very messy, but my final answer was:

1/(s+2) - 1/((s+2)^2) - 1/((s+2)^3) , s =/= -2

is this right?

Printable View

- Apr 17th 2011, 07:26 PMlinalg123la place transform of t^2e^(-2t) from definition
if i try to write out what i did on this it will be very messy, but my final answer was:

1/(s+2) - 1/((s+2)^2) - 1/((s+2)^3) , s =/= -2

is this right? - Apr 17th 2011, 07:57 PMTKHunny
You should eliminate the mess and somehow lose the first two pieces.

Do t^2 and e^(-2t) separately and get an idea of where you are aiming. - Apr 17th 2011, 08:40 PMProve It
If you are using the definition, you should know that

http://quicklatex.com/cache3/ql_27fe...80060a6_l3.png.

It's not pretty, but here you go...

http://quicklatex.com/cache3/ql_f45e...31dedc2_l3.png.

Of course, it's much, much, much easier to use the fact that http://quicklatex.com/cache3/ql_9549...cfef953_l3.png, because then...

http://quicklatex.com/cache3/ql_5094...ef122f8_l3.png. - Apr 18th 2011, 01:36 AMlinalg123
thanks for that.. one question..

why do you differenciate the first term only in the 6th line, and why is that acceptable?

i.e when lim {ε^2/e^((s+2)ε)} goes to lim {2ε/[(2+s)e^(2+s)ε]} - Apr 18th 2011, 04:07 AMProve It
I didn't, I used L'Hospital's Rule.

- Apr 18th 2011, 05:55 PMlinalg123
ahhhh yes i see... thanks heaps