2nd Order differential equation using laplace

**simpo067** · April 12th 2008, 05:38 AM

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

**Peritus** · April 12th 2008, 06:12 AM

$ 2s^2 + 6s + 4.5 = 2(s + 1.5)^2 $

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

a known transform pair is:

$ 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:

$ L^{ - 1} \left\{ {\frac{4} {{(s + 1.5)^2 }}} \right\} = 4te^{ - 1.5t} u(t) $

**simpo067** · April 12th 2008, 06:17 AM

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.

Thread: 2nd Order differential equation using laplace

LinkBack

Thread Tools

Display

2nd Order differential equation using laplace

Similar Math Help Forum Discussions

Basic Laplace Transform of a 1st order Differential Equation

Laplace Transform of a 2nd order Differential Equation

Solve second order differential equation by Laplace transform

First-order differential equation (Laplace)

Higher-order differential equations, the Laplace transform and exponential order

Search Tags