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I've gotten to \[2^2L(y)-6-y=5*\sin2t\]

but I think I'm stuck... that should say s^2 not 2^2

\[s^2L(y)-6-y=5*\sin(2t)\]

do you have the value of y'(0)?

yep y'(0)=6

I messed that all up.... it should be\[s^2L(y)-6-L(y)=10/s^2+4\]

yeah this is right

Ok so what do I do from there?

try to solve for L(y)

I thikn I would go \[s^2(L(y)-1)-6=10/s^2+4\]

err

ok I prefer to write L(y) as Y(s)

\[L(y)(s^2-1)-6=10/s^2+4\]

it should be :
\[Y(s)(s^2-1)={10 \over s^2+4} +6\]

then you divide each side by s^2-1 and then look up the transform on a table?

=>6s^2+34 = Y{s^2 - 1}

Im confused on where you got that thinker

take the LCM :)

= \[Y(s)={6s^2+34 \over s^2+4} . {1 \over s^2 -1}\]

sry :) missed the denominator :)

are you follwoing Scotty?

now i guess u may use partial fractions to solve further

1 sec I think so im writing it out

take your time.. I will be right back

try a way to write it back in time domain

Yep I got it, you were adding 6 into there.

Can I look up the laplace transform inverse for each and then combine them?

Forgot how to do those, you're talking about breaking it up right?

take , say, (A/s^ + 4)+(b/s^2 -1)

sry it is (A/s^2 + 4)+(b/s^2 -1)

mm ok
I got it now. Thanks guys... I'm sure I'm gonna post another one here in a min lol

I am back

I think there is something wrong

no I just ran what we had through my solver and it came up with the correct answer...

I mean the partial fractions

whats that?? u know it was 2 years back i studies Laplace transforms :):)

:)

Oh well the laplace transform before you get to the fractions are correct :)

^^
yeah I know :)

omg!! did i forget solving these?? lol

right

was my suggestion of no help ??

No you helped thinker. Thank you :)

it doesn't matter what notation you use as long as you understand it.