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Laplace Transform

20140424_154836.jpg20140424_154858.jpgQ: y"-y'-6y=cos t y (0)=1 and y '(0)=2

The question only wants the solution by using the laplace transform.

I did all I could, I don't have the answer so I go to wolfram to verify and I got it close. The answer from wolfram is y=c1 e^-2t + c2 e^3t - sin t /50 - 7cos t /50

but my answer is -e^t/2 + 53e^6t/35 - 29sin t /70 - cos t/70
(edited 9 years ago)
Original post by Taffss
20140424_154836.jpg20140424_154858.jpgQ: y"-y'-6y=cos t y (0)=1 and y '(0)=2

The question only wants the solution by using the laplace transform.

I did all I could, I don't have the answer so I go to wolfram to verify and I got it close. The answer from wolfram is y=c1 e^-2t + c2 e^3t - sin t /50 - 7cos t /50

but my answer is -e^t/2 + 53e^6t/35 - 29sin t /70 - cos t/70


Your partial fractions are up the creek.

You had (s^2-s-6) in the denominator at one point.

Which factorises to.... (s1)(s6)\not= (s-1)(s-6)
Reply 2
Original post by ghostwalker
Your partial fractions are up the creek.

You had (s^2-s-6) in the denominator at one point.

Which factorises to.... (s1)(s6)\not= (s-1)(s-6)


Oh my God how can I do that. I feel foolish. Thank you

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