Laplace DE with unit step function
Just a quick question on a laplace transform question,
I need to find the transform for
y" +2y' + y = u(t-1) , y(0)=1, y'(0)= -1
After using the known properties i obtained,
Y(s) = exp(-s)[1/(s(s-1)^2)] + (s^2+s)/(s(s-1)^2)
using partial fractions, t-shifting and s-shifting and other known properties i arrived at
y(t) = (1+exp(t-1)((t-1)-1))u(t-1) + exp(t)(1+t)
Thus for 0<t<1
y(t) =exp(t)(1+t)
for t>1
y(t)= 1+exp(t-1)((t-1)-1) + exp(t)(1+t)
Is my answer and methodology correct? There is no answer in the book.
Thanks for your help.
Re: Laplace DE with unit step function
I think you might have some sign errors in there somewhere. Taking the LT of the original DE yields
)-sy(0)-y'(0)=\frac{e^{-s}}{s},)
which leads to
=\frac{e^{-s}}{s}+s+1.)
I end up with the LT being
^{2}}+\frac{1}{s+1}.)
You can use partial fractions on the first to obtain
![Y=e^{-s}\left[\frac{1}{s}-\frac{1}{s+1}-\frac{1}{(s+1)^{2}}\right]+\frac{1}{s+1}.](http://latex.codecogs.com/png.latex?Y=e^{-s}\left[\frac{1}{s}-\frac{1}{s+1}-\frac{1}{(s+1)^{2}}\right]+\frac{1}{s+1}.)
Can you finish from here?
