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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 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post szz Differential Equations 1 November 2nd, 2014 03:18 AM Bkalma Calculus 1 May 6th, 2014 07:08 PM Asmir Complex Analysis 2 August 18th, 2013 05:49 AM capea Complex Analysis 5 August 3rd, 2013 02:30 PM Deiota Calculus 1 April 28th, 2013 10:28 AM

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