[QUOTE="CHOASXIII"][QUOTE="ghoklebutter"]
It's pretty nice actually - here's a typical example:
y'' + 5y' + 6y = e^x
[D^2]y + 5Dy + 6y = e^x
[D^2 + 5D + 6](y) = e^x
Because [D - 1]e^x = De^x - e^x = e^x - e^x = 0
Multiply each side by [D - 1] to annihilate the right side of the DE:
[D - 1][D^2 + 5D + 6](y) = [D - 1][D + 2][D + 3](y) = 0
D = -3, -2, 1, where D = 1 is a root for the nonhomogenous solution and D = -3, -2 are roots for the homogenous one.
The complete solution is y(x) = ae^(-3x) + be^(-2x) + ce^x
The only real downside is that it only works for nonhomogenous equations that involve exponential, polynomial, and/or trigonometric functions on the right side.
SaudiFury
Get that moonspeak off my OT.
lol.. at least i got a refresher on laplace transformations. haven't seen it in a while.I don't even know what any of that is.
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