I'm trying to solve the following differential equation:(adsbygoogle = window.adsbygoogle || []).push({});

y'' + 2y' + y = 2t^2 - 1

y(0) = 1

y'(0) = 0

I'm trying to figure this out by looking at the solution to a similar diff. eq.

y' + 2y = t + 1

y(0) = 1

First step is to set the left part equal to 0

y' + 2y = 0

Then do a laplace transform on both sides:

L(y' + 2y) = L(0)

sY + 2Y = 0

solving the above gives s = -2

plug that into y = Ae^(st) to get y = A^(-2t)

If I were to go this far with the first problem, what would it look like?

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# Differential equation

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