Laplace Transform/Unit Step Func.

**Auxiliary** · December 10th 2007, 01:21 PM

Solve the IVP:

$\frac{1}{4}x'(t) + 4x(t) = \frac{t}{2} - \frac{t}{2}\mu(t-10),\,x(0) = 8$

where

$\mu(t-a) = \begin{cases} 0,& \text{if $0\leq t< a$}\\ 1,& \text{if $t \geq a$}\end{cases}$

NOTE: I used $\mu$ because I wasn't sure how to do the capital curly "U" that the book uses. This is called the "unit step function".

**Auxiliary** · December 12th 2007, 12:04 PM

Does anyone know how to do laplace transforms?

. I keep trying it, and I keep getting different results. Here are my attempts:

NOTE: I will denote L to be the laplace operator symbol as I don't know how to do a curly L, and [ and ] will denote { and }, respectively.

$L[\frac{1}{4}\frac{dx}{dt}] + L[4x(t)] = L[\frac{t}{2}] - L[\frac{t}{2}\mu(t-10)]$

$L[\frac{dx}{4dt}] = sX(s) - 8$

$L[4x(t)] = 4X(s)$

$L[\frac{t}{2}] = \frac{1}{2s^2}$

$L[\frac{t}{2}\mu(t-10)] = e^{-10s}L[\frac{t+10}{2}] = e^{-10s}\left(\frac{1}{2s^2}+\frac{5}{s}\right)$

$sX(s) - 8 + 4X(s) = \frac{1}{2s^2} - \frac{e^{-10s}}{2s^2} - \frac{5e^{-10s}}{s}$

$X(s)(s+4) = \frac{1}{2s^2} - \frac{e^{-10s}}{2s^2} - \frac{5e^{-10s}}{s} + 8$

Use partial fractions:

$L[X(s)] = L[\left(\frac{-1}{32s} + \frac{1}{32(s+4)} + \frac{1}{8s^2}\right) - \mu(t-10)\left(\frac{39}{325} - \frac{39}{32(s+4)} + \frac{1}{8s^2}\right) + \frac{8}{s+4}]$

$L[X(s)] = x(t)$

$X(t) = -\frac{1}{32} + \frac{1}{32}e^{-4t} + \frac{1}{8}t - \mu(t-10)\left(\frac{39}{32} - \frac{39}{32}e^{-4(t-10)} + \frac{1}{8}(t-10)\right) + 8e^{-4t}$

And then not sure how to get this to piecewise func, but I don't think this is right any way. Please help!

**Auxiliary** · December 12th 2007, 02:50 PM

No one knows of Laplace Transforms?

I attached a file to show an example done in-class ... this problem is frustrating me. It's the last thing I have to do in this course too!

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