Laplace shifting problem

**weedingb** · October 28th 2010, 12:26 AM

Hi,

I am trying to solve

$y''+4y=0, 0<t<a; exp(-t), t>a$

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

**Ackbeet** · October 28th 2010, 02:19 AM

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?

**chisigma** · October 28th 2010, 05:08 AM

Originally Posted by weedingb

Hi,

I am trying to solve

$y''+4y=0, 0<t<a; exp(-t), t>a$

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

$\chi$ $\sigma$

**chisigma** · October 28th 2010, 12:44 PM

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

$\chi$ $\sigma$

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