Solve differential equation using inverse laplace transform: y''-4y'+3y=6t-8, y(0)=y'(0)=0.

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Solve differential equation using inverse laplace transform: y''-4y'+3y=6t-8, y(0)=y'(0)=0.

Mathematics
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\[Y=(6-8s)\div((s-3)(s-1)s^2)\] is what I have so far, I know I need to do partial fraction decomp and the integral because of the s^2 in the denominator. Does it matter which I do first?
there's no integration involved, you simply have to use partial fractions so it is in a more appropriate form to inverse laplace
Don't you have to do partial fractions to solve (6-8s)/((s-3)(s-1), and then something with an integral to deal with the s^2 in the denominator? Partial fractions doesn't give me a complete answer. I can get e^t-e^3t, but the answer also includes a 2t which I'm not getting

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Nevermind, I got it. You were right.
Yeah, laplace is pretty much always the same process so if you get that down it's just algebra after that

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