# Laplace DE with unit step function

• Sep 19th 2011, 05:53 PM
olski1
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.

• Sep 20th 2011, 06:44 PM
Ackbeet
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

$\displaystyle s^{2}Y+Y+2(sY-y(0))-sy(0)-y'(0)=\frac{e^{-s}}{s},$ which leads to

$\displaystyle Y(s^{2}+2s+1)=\frac{e^{-s}}{s}+s+1.$

I end up with the LT being

$\displaystyle Y=\frac{e^{-s}}{s(s+1)^{2}}+\frac{1}{s+1}.$

You can use partial fractions on the first to obtain

$\displaystyle Y=e^{-s}\left[\frac{1}{s}-\frac{1}{s+1}-\frac{1}{(s+1)^{2}}\right]+\frac{1}{s+1}.$

Can you finish from here?