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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- anonymous

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

yep y'(0)=6

- anonymous

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

- anonymous

yeah this is right

- anonymous

Ok so what do I do from there?

- anonymous

try to solve for L(y)

- anonymous

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

- anonymous

err

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Im confused on where you got that thinker

- anonymous

take the LCM :)

- anonymous

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

- anonymous

sry :) missed the denominator :)

- anonymous

are you follwoing Scotty?

- anonymous

now i guess u may use partial fractions to solve further

- anonymous

1 sec I think so im writing it out

- anonymous

take your time.. I will be right back

- anonymous

try a way to write it back in time domain

- anonymous

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

- anonymous

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

- anonymous

i guess its possible when there's addition there..here we have multiplication of terms...can u use partial fractions?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I am back

- anonymous

I think there is something wrong

- anonymous

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

- anonymous

I mean the partial fractions

- anonymous

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

- anonymous

:)

- anonymous

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

- anonymous

^^
yeah I know :)

- anonymous

omg!! did i forget solving these?? lol

- 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

right

- 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

I didn't learn the notation in form of Y(s) which seems to be the way most people learn it. Very annoying.

- anonymous

was my suggestion of no help ??

- anonymous

No you helped thinker. Thank you :)

- anonymous

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

- anonymous

thnx :) @scott

Looking for something else?

Not the answer you are looking for? Search for more explanations.