# DE's with discontinuous forcing functions

• Dec 14th 2008, 05:14 PM
petition Edgecombe
DE's with discontinuous forcing functions
I'm having trouble with a bunch of these kinds of problems:

$y''+4y=sin(t)-u_{2\pi}(t)sin(t-2\pi); \;\;\; y(0)=0, \; y'(0)=0$

So far I've got:

$s^2Y(s)-sy(0)-y'(0)+4Y(s)-y(0)=\frac{1}{s^2+1}-\frac{e^{-2\pi}}{s^2+1}$

$\Rightarrow Y(s)=\frac{1}{(s^2+1)(s^2+4)}-\frac{e^{-2\pi}}{(s^2+1)(s^2+4)}$

And I'm laboring with the partial fraction decomposition:

$1=(As+B)(s^2+4)+(Cs+D)(s^2+1)$
• Dec 14th 2008, 06:15 PM
Krizalid
Observe that $(s^2+4)-(s^2+1)=3.$

Pretty sure that someone else's gonna provide more help, 'cause now I'm going to sleep. :)
• Dec 15th 2008, 02:01 AM
petition Edgecombe
Still don't get it.
• Dec 15th 2008, 02:16 AM
Mush
Quote:

Originally Posted by petition Edgecombe
I'm having trouble with a bunch of these kinds of problems:

$y''+4y=sin(t)-u_{2\pi}(t)sin(t-2\pi); \;\;\; y(0)=0, \; y'(0)=0$

So far I've got:

$s^2Y(s)-sy(0)-y'(0)+4Y(s)-y(0)=\frac{1}{s^2+1}-\frac{e^{-2\pi}}{s^2+1}$

$\Rightarrow Y(s)=\frac{1}{(s^2+1)(s^2+4)}-\frac{e^{-2\pi}}{(s^2+1)(s^2+4)}$

And I'm laboring with the partial fraction decomposition:

$1=(As+B)(s^2+4)+(Cs+D)(s^2+1)$

$(s^2+4) - (s^2+1) = 3$

So if you multiply top and bottom of both terms by 3:

$\Rightarrow Y(s)=\frac{3}{3(s^2+1)(s^2+4)}-\frac{3e^{-2\pi}}{3(s^2+1)(s^2+4)}$

$\Rightarrow Y(s)=\frac{(s^2+4) - (s^2+1)}{3(s^2+1)(s^2+4)}-\frac{((s^2+4) - (s^2+1))e^{-2\pi}}{3(s^2+1)(s^2+4)}$

Now split up each fraction into 2 terms

$\Rightarrow \frac{s^2+4}{3(s^2+1)(s^2+4)} - \frac{s^2+1}{3((s^2+1)(s^2+4)} - \frac{(s^2+4)e^{-2 \pi}}{3(s^2+1)(s^2+4)} + \frac{(s^2+1)e^{-2 \pi}}{3(s^2+1)(s^2+4)}$

Easier now?
• Dec 15th 2008, 03:53 AM
petition Edgecombe
But what is the Inverse Laplace transform of $\frac{1}{s^2+4}$ ?
• Dec 15th 2008, 04:05 AM
Mush
Quote:

Originally Posted by petition Edgecombe
But what is the Inverse Laplace transform of $\frac{1}{s^2+4}$ ?

$\mathcal L (\frac{k}{s^2+a^2}) = k. \frac{sin(at)}{a}$

In this case a = 2