Another laplace differential equation.

  • Thread starter TyErd
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  • #1
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Homework Statement


question is attached

Homework Equations


L(1) = 1 /s
L(t^n) = n!/s^(n+1)
L{f '(t)} = s F(s) - f(0)
L{f ''(t)} = s^2 F(s) - s f(0) - f '(0)

The Attempt at a Solution



okay so i laplace transformed both sides and i got:

(s^2) Y(s) + 5s -5 -5s Y(s) -25 - 6Y(s) = -84exp(-5s) / s
Y(s) [s^2 - 5s - 6] + 5s - 30 = -84exp(-5s) / s
Y(s) = ((-exp(-5s)(5s(s-6)*exp(5s) +84)) / (s^3 - 5s^2 - 6s)

but thats so long and complicated that im wondering if i've stuffed up because I can't even seem to find the inverse laplace of the function.
 

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Answers and Replies

  • #2
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In your first line, I think I found a problem:

(s^2) Y(s) + 5s -5 -5s Y(s) -25 - 6Y(s) = -84exp(-5s) / s

The laplace transform of the forcing function -84H(t-5), assuming H is the unit step function, that laplace transform should be different.

The table I have claims that:
[tex]L{f(t)U(t-a)) = e^{-as}L(F(t+a))[/tex]

Since yout f(t) is basically 1, if you just pull out the constant -84. the laplace of which is 1/s, so the laplace of f(t+a) should be 1/(s+a) I believe.
 
  • #3
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laplace of f(t+a) isnt 1/(s+a) i think.
 
  • #4
1,045
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I'm looking at the table right now.

The laplace transform of f(t)U(t-1) is e^(-as) times the laplace transform of f(t+a).

Oh wait, I see what you mean.
 
  • #5
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okay so that would mean laplace of -84 H(t-5) = -84exp(-5s) / (s+5) yea?
 
  • #6
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That's how I interpret it.
 
  • #7
299
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so my working out now is: (s^2) Y(s) + 5s -5 -5s Y(s) -25 - 6Y(s) = -84exp(-5s) / s+5

therefore Y(s) = -exp(-5s)(5(s-6)(s+5)*exp(5s) +84) / (s-6)(s+1)(s+5)
making y(t) = -1/11exp(-5(t+6))*[3(4exp(11t) - 11exp(4t+35) + 7exp(55) H(t-5)+55 exp(4t+30)] which is way to long to be right.
 

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