# 2nd Order differential equation using laplace

• Apr 12th 2008, 05:38 AM
simpo067
2nd Order differential equation using laplace
Banging my head against a wall so some help would be appreciated. I've drilled a second order differential using laplace down to:

8/(2s^2+6s+4.5).

I just can't seem to figure out the inverse laplace transform past the initial stage of taking out the numerator and dividing the denominator through by 2. Help much appreciated. Should be able to see this but just can't
• Apr 12th 2008, 06:12 AM
Peritus
$\displaystyle 2s^2 + 6s + 4.5 = 2(s + 1.5)^2$

$\displaystyle \Leftrightarrow \frac{8} {{2s^2 + 6s + 4.5}} = \frac{4} {{(s + 1.5)^2 }}$

a known transform pair is:

$\displaystyle L\left\{ {\frac{{t^n }} {{n!}}u(t)} \right\} = \frac{1} {{s^{n + 1} }}$

now using the Frequency shifting property of the Laplace transform we can deduce that:

$\displaystyle L^{ - 1} \left\{ {\frac{4} {{(s + 1.5)^2 }}} \right\} = 4te^{ - 1.5t} u(t)$
• Apr 12th 2008, 06:17 AM
simpo067
Thanks for that

I was just about to post a "don't bother the clouds have just parted message". Don't know why i couldn't factorise the damn thing. Thanks again.

Final answer i came up with was was 4e^-1.5t * t (from n!/(s-a)^n+1 = e^at*t^n ) as this was listed in my maths book. Thanks again and can i ask what editor you used to enter the reply? as I can't seem to cut and paste from mathtype or equation editor.