
Differential Equations Ordinary and Partial Differential Equations Math Forum 
 LinkBack  Thread Tools  Display Modes 
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  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Relatioship between Laplace and Inverse Laplace Transform from tables  szz  Differential Equations  1  November 2nd, 2014 03:18 AM 
Inverse Laplace Transform  Bkalma  Calculus  1  May 6th, 2014 07:08 PM 
Inverse Laplace transform  Asmir  Complex Analysis  2  August 18th, 2013 05:49 AM 
inverse laplace transform  capea  Complex Analysis  5  August 3rd, 2013 02:30 PM 
Laplace tranform and inverse of Laplace transform  Deiota  Calculus  1  April 28th, 2013 10:28 AM 