My Math Forum Inverse Laplace Transform of a 2nd Order ODE

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 April 18th, 2015, 01:25 PM #1 Newbie   Joined: Apr 2015 From: somewhere Posts: 1 Thanks: 0 Inverse Laplace Transform of a 2nd Order ODE 23.) $\displaystyle y'' +2y' + y = 4 e^-t; y(0) = 2, y'(0) = -1$ My attempt: $\displaystyle a = 1, b = 2, c = 1$ $\displaystyle F(s) = 4 L( e ^ -t) = 4/(s + 1)$ (Taken from Laplace Transform Table) $\displaystyle Y(s) = [(as + b) y(0) + a y'(0) + F(s)]/(as^2 + bs + c)$ Plugging and Simplifying: $\displaystyle Y(s) = (2s^2 + 5s + 7)/[(s + 1)(s^2 + 2s + 1)]$ This is where I'm stuck. I find it peculiar that the denominator is a perfect cube. I've hit it with partial fraction decompositon, but my attempts haven't worked well, and the numerator doesn't factor. Any help would be awesome. Thanks! Post note: Sorry my formatting is terrible. The exponential is raised to the - t. I cannot seem to get it to work.
 April 18th, 2015, 11:47 PM #2 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Your question is illegible: each equation appears as : [Math Processing Error]

 Tags 2nd, inverse, laplace, ode, order, transform

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