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Leplace transform

  • Thread starter hils0005
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  • #1
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Homework Statement



y"+4y=7t , y(0)=-1, y'(0)=3


The Attempt at a Solution



L{y} + 4L{y} = 7L{t}

(s^2L{y} - sy(0) - y'(0)) + 4L{y} = 7L{t}

s^2L{y} - s(-1) - 3 + 4L{y} = 7(0!/s^(0+1))

L{y}(s^2 + 4) = 7(0!/s^(0+1)) - s + 3
= ((-s+3)s + 7)/s
=(-s^2 + 3s +7)/s

L{y}=(-s^2 + 3s + 7)/(s(s^2+4))

here is where I run into trouble, setting up the partial fractions

L{y}=-s^2 + 3s +7 = A/s + B/(s^2+4)

is that correct so far?
 

Answers and Replies

  • #2
15
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You should check the laplace transform for 7t.

L{t^n} = (n!)/(s^(n+1))
therefore L{t} = L{t^1} = (1!/(s^(1+1))

L{7t} should look more like 7*(1/s^2) or (7/s^2)
 
  • #3
62
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but its t^1, so n=0, t^0+1=t^1 right?
 
  • #4
15
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Its t^1, so n = 1 when you plug into the transform for t^n.
 
  • #5
1,631
4
Or if you want the Laplace transform of t^n in terms of gamma function:

[tex] L\{t^n\}=\frac{\Gamma(n+1)}{s^{n+1}}[/tex]

[tex]\Gamma(n+1)=n\Gamma(n)=.....=n!,n \in Z^+[/tex]
[tex]\Gamma(1)=1[/tex]
 
Last edited:

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