Need help with LaPlace transform y''-y=5*sin2t when given y(0)=0, y'(0)=6

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Need help with LaPlace transform y''-y=5*sin2t when given y(0)=0, y'(0)=6

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

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Other answers:

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?
i guess its possible when there's addition there..here we have multiplication of terms...can u use partial fractions?
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
if that's all you want, then it's ok.. but if you want to fin the final solution for the differential equation by yourself, then you need to do the partial fractions
right
the point is that you want to write the expression we have for Y(s) in a form that is easy to be transformed back to time domain
I didn't learn the notation in form of Y(s) which seems to be the way most people learn it. Very annoying.
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.
thnx :) @scott

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