Find the Laplace Transformation of y'' - 3y' + 2y = 4e^{-2x} when y(0) = 1, y'(0) = 4 Y(s) = (s^2 + 4s - 6) / (s^3 - s^2 - 4s + 4) Y(s) = (s^2 + 3s - 6) / (s^3 - s^2 + 4s + 4) Y(s) = (s^2 - 4s + 8) / (s^3 - s^2 - 4s + 4) Y(s) = (s^2 + 3s + 6) / (s^3 - s^2 - 4s + 4) Y(s) = (s^2 + 3s + 8) / (s^3 - s^2 + 4s + 4) Y(s) = (s^2 + s + 8) / (s^3 - s^2 + 4s + 4) Y(s) = (s^2 - 3s + 8) / (s^3 - s^2 - 4s + 4) Y(s) = (s^2 + 2s + 6) / (s^3 - s^2 - 4s + 4)
Added by Michael R.
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The given differential equation is y - 3y' + 2y'' = 4e^(-2x). Show more…
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