ODE with Laplace Transform: Solving for Y(s) and Partial Fraction Expansion

bosox09
Messages
7
Reaction score
0

Homework Statement



The solution to the ODE y''(t) + 4y(t) = 1 + u(t − 2), y(0) = 0, y'(0) = 0 is given by...

The Attempt at a Solution



OK well I figured this one is good to solve with Laplace transforms. So I took the Laplace of both sides to obtain (s2 + 4)Y(s) = [e-2s/s] + 1/s, which equals (e-2s + 1)/s. Isolating Y(s) gave me (e-2s + 1)/s(s2 + 4). I used partial fraction expansion to obtain (1/4) - (1/4)cos2t, but this is apparently only half of the whole answer, given as (1/4)(1 − cos 2t) + (1/4)(1 − cos 2(t − 2))u(t − 2). What am I missing?
 
Physics news on Phys.org
How are you expanding in partial fractions? What expansion did you get? In particular shat happened to the e^{-2s}? It's very easy to see that y= \frac{1}{4}(1- cos(2t)) satisfies y"+ 4y= 1, y(0)= 0, y'(0)= 0, not the equation you have.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top