# Laplace shifting problem

• Oct 28th 2010, 12:26 AM
weedingb
Laplace shifting problem
Hi,

I am trying to solve

$y''+4y=0, 0a$

using Laplace transforms.

I end up with
$H(t-a)(exp(-(t-a))-cos(2(t-a))+1/2sin(2(t-a))$

where H(t-a) is the Heaviside/unit step function. Is this correct? If not, where have I gone wrong?

Cheers,
Ben
• Oct 28th 2010, 02:19 AM
Ackbeet
Hmm. That isn't what I get. You're missing the complimentary function, I think.

1. How can you write the original DE in one equation using the Heaviside step function?
2. What do you get when you LT the DE?
3. What are the initial conditions?
4. So what do you get when you solve for the LT of y?
• Oct 28th 2010, 05:08 AM
chisigma
Quote:

Originally Posted by weedingb
Hi,

I am trying to solve

$y''+4y=0, 0a$

using Laplace transforms.

I end up with
$H(t-a)(exp(-(t-a))-cos(2(t-a))+1/2sin(2(t-a))$

where H(t-a) is the Heaviside/unit step function. Is this correct? If not, where have I gone wrong?

Cheers,
Ben

You have proposed a second order linear DE for the solution of which it is necessary to know the values $y(0)$ and $y^{'}(0)$ at the moment not yet specified...

Kind regards

$\chi$ $\sigma$
• Oct 28th 2010, 12:44 PM
chisigma
The DE is [in my opinion]...

$y^{''} + 4\ y= \left\{\begin{array}{ll} 0,\,\,0 < x < a{}\\e^{-t} ,\,\, x\ge a\end{array}\right.$ (1)

... and if we suppose $y(0)= y^{'}(0)=0$ in terms of Laplace Transform is written...

$\displaystyle s^{2}\ Y(s) + 4\ Y(s) = \frac{e^{-a\ (s+1)}}{s+1} \implies Y(s)= \frac{e^{-a\ (s+1)}}{(s+1)\ (s^{2}+4)}$ (2)

Now if You consider that is...

$\displaystyle \frac{1}{(s+1)\ (s^{2}+4)} = \frac{1}{5}\ \{\frac{1}{s+1}+ \frac{1-s}{s^{2}+4} \}$ (3)

... You obtain...

$\displaystyle y(t)= \mathcal{L}^{-1} \{Y(s)\} = \frac{e^{-a}}{5}\ \{e^{-(t-a)} + \cos 2(t-a) - \sin 2(t-a)\}\ \mathcal{H} (t-a)$ (4)

Kind regards

$\chi$ $\sigma$