Initial value problem (Laplace)

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  • #1
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Homework Statement



Use the Laplace transform method to solve the differential equation

y" -4y' +4y = e^2t subject to initial conditions y(0)=1, y'(0)=0

Homework Equations





The Attempt at a Solution



s^2Y(s) -sy(0) - y'(0) - 4[ sY(s) - y(0)] + 4Y(s)= L (e^2t)

Y(s) [s^2 -4s +4] - s -4 = 1/(s-2)

Y(s) [ (s-2)^2 ] = 1/(s-2) + s +4

Y(s) = 1/(s-2)^3 + s/(s-2)^2 + 4/(s-2)^2

Is this right, how do I find out the inverse laplace of Y(s) like for example 1/(s-2)^3 seems impossible!

Thanks.
 

Answers and Replies

  • #2
HallsofIvy
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??? Pretty nearly every inverse lapalce transform is done by looking it up in a table of transforms. But, as I pointed out in response to your other post, generally, you find the Laplace transform (and almost always the inverse transform) by looking it up in a table. Here is a good one:
http://www.vibrationdata.com/Laplace.htm

In particular, the inverse transform of 1/(s-2)3 is (1/2)t2e-2t.
 
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  • #3
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how about s/(s-2)^2. It isnt a cosine or sine laplace thing?

Thanks.
 
  • #4
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You can find [tex]\frac{1}{(s-2)^2}[/tex] and s in that table he linked.
 
  • #5
HallsofIvy
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[tex]\frac{1}{(s-2)^3}[/tex]
and
[tex]\frac{1}{(s-2)^2}[/tex]
are given by number 2.11 in that table.

By partial fractions,
[tex]\frac{s}{(s-2)^2}= \frac{1}{s-2}+ \frac{2}{(s-2)^2}[/tex]
 

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