Can't seem to get this DE solved using laplace transforms

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SUMMARY

The discussion centers on solving the differential equation y'' - 4y' + 4y = 0 using Laplace transforms. The user initially struggles with the transformation process but ultimately identifies an error in their numerator, realizing it should be s - 3 instead of s + 3. The correct solution after applying the inverse transform is (5t + 1)e^(2t), confirming the importance of accurate sign management in Laplace transforms.

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Homework Statement



y''-4y'+4y = 0
y(0)=1, y'(0)=1

Homework Equations





The Attempt at a Solution



This is annoying, because it feels very easy and I don't know why I am not getting the answer. Skipping the routine:

s^2L{y} - s - 1 - 4sL{y} + 4 + 4L{y} = 0

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

= s/(s-2)^2 + 3/(s-2)^2

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

= 5(1/(s-2)^2) + 1/(s-2)


The inverse transform of which gives:

5te^2t + e^2t = (5t+1)e^2t


Any ideas?

Thanks!
 
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NVM, I think it should be s-3 in num now.
 
Check your signs.

EDIT: Ah, I see you found the mistake.
 

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