Laplace Transform

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  • #1
tony873004
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From the class notes:
[tex]\begin{array}{l}
y'' + 8y' + 16y = te^{ - 4t} ,\,\,\,\,\,y\left( 0 \right) = y'\left( 0 \right) = 0 \\
\\
L\left[ {y''} \right] + 8L\left[ {y'} \right] + 16L\left[ y \right] = \frac{1}{{\left( {s + 4} \right)^2 }} \\
\end{array}[/tex]

How did he get [tex]\frac{1}{{\left( {s + 4} \right)^2 }}[/tex] ?
From the table, [tex]t = \frac{1}{{s^2 }}[/tex] and [tex]e^{at} \to \frac{1}{{s - a}}[/tex]
How do these combine to give [tex]\frac{1}{{\left( {s + 4} \right)^2 }}[/tex] ?

The next line is
[tex]s^2 y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) + 8\left( {sy\left( s \right) - y\left( 0 \right) + 16y\left( s \right)} \right) = \frac{1}{{\left( {s + 4} \right)^2 }}[/tex]

Where did everything on the left side of = come from? The table doesn’t have y’’ or y’.

After this, the problem looks like it turns into algebra.
 

Answers and Replies

  • #2
lanedance
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Hi Tony

The laplace transform of a product is not just the Laplace transform of the components, have a look at this table:
http://www.efunda.com/math/laplace_transform/forward.cfm?FuncName=Basic
shows:
[tex]L(e^{-\alpha t}) = \frac{1}{{\left( {s + 4} \right)^2 }}[/tex]
to get the relation you actually need to perform the integral


the next line comes about from the Laplace transform rules for derivatives, see this table
http://www.vibrationdata.com/Laplace.htm
these can be derived using integration by parts on successive derivatives if i remember rightly...
 
  • #3
tony873004
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Thanks. The 6th entry in the 1st table you linked to has the right side of my equation. It was missing from the table I had from class notes. And thanks for the 2nd table. I think it explains it.
 

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