anonymous
  • anonymous
Need help with LaPlace transform y''-y=5*sin2t when given y(0)=0, y'(0)=6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I've gotten to \[2^2L(y)-6-y=5*\sin2t\]
anonymous
  • anonymous
but I think I'm stuck... that should say s^2 not 2^2
anonymous
  • anonymous
\[s^2L(y)-6-y=5*\sin(2t)\]

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anonymous
  • anonymous
do you have the value of y'(0)?
anonymous
  • anonymous
yep y'(0)=6
anonymous
  • anonymous
I messed that all up.... it should be\[s^2L(y)-6-L(y)=10/s^2+4\]
anonymous
  • anonymous
yeah this is right
anonymous
  • anonymous
Ok so what do I do from there?
anonymous
  • anonymous
try to solve for L(y)
anonymous
  • anonymous
I thikn I would go \[s^2(L(y)-1)-6=10/s^2+4\]
anonymous
  • anonymous
err
anonymous
  • anonymous
ok I prefer to write L(y) as Y(s)
anonymous
  • anonymous
\[L(y)(s^2-1)-6=10/s^2+4\]
anonymous
  • anonymous
it should be : \[Y(s)(s^2-1)={10 \over s^2+4} +6\]
anonymous
  • anonymous
then you divide each side by s^2-1 and then look up the transform on a table?
anonymous
  • anonymous
=>6s^2+34 = Y{s^2 - 1}
anonymous
  • anonymous
Im confused on where you got that thinker
anonymous
  • anonymous
take the LCM :)
anonymous
  • anonymous
= \[Y(s)={6s^2+34 \over s^2+4} . {1 \over s^2 -1}\]
anonymous
  • anonymous
sry :) missed the denominator :)
anonymous
  • anonymous
are you follwoing Scotty?
anonymous
  • anonymous
now i guess u may use partial fractions to solve further
anonymous
  • anonymous
1 sec I think so im writing it out
anonymous
  • anonymous
take your time.. I will be right back
anonymous
  • anonymous
try a way to write it back in time domain
anonymous
  • anonymous
Yep I got it, you were adding 6 into there.
anonymous
  • anonymous
Can I look up the laplace transform inverse for each and then combine them?
anonymous
  • anonymous
i guess its possible when there's addition there..here we have multiplication of terms...can u use partial fractions?
anonymous
  • anonymous
Forgot how to do those, you're talking about breaking it up right?
anonymous
  • anonymous
take , say, (A/s^ + 4)+(b/s^2 -1)
anonymous
  • anonymous
sry it is (A/s^2 + 4)+(b/s^2 -1)
anonymous
  • anonymous
mm ok I got it now. Thanks guys... I'm sure I'm gonna post another one here in a min lol
anonymous
  • anonymous
I am back
anonymous
  • anonymous
I think there is something wrong
anonymous
  • anonymous
no I just ran what we had through my solver and it came up with the correct answer...
anonymous
  • anonymous
I mean the partial fractions
anonymous
  • anonymous
whats that?? u know it was 2 years back i studies Laplace transforms :):)
anonymous
  • anonymous
:)
anonymous
  • anonymous
Oh well the laplace transform before you get to the fractions are correct :)
anonymous
  • anonymous
^^ yeah I know :)
anonymous
  • anonymous
omg!! did i forget solving these?? lol
anonymous
  • anonymous
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
anonymous
  • anonymous
right
anonymous
  • anonymous
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
anonymous
  • anonymous
I didn't learn the notation in form of Y(s) which seems to be the way most people learn it. Very annoying.
anonymous
  • anonymous
was my suggestion of no help ??
anonymous
  • anonymous
No you helped thinker. Thank you :)
anonymous
  • anonymous
it doesn't matter what notation you use as long as you understand it.
anonymous
  • anonymous
thnx :) @scott

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