Laplace Transform

tony873004

Science Advisor
Gold Member
1,748
141
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.
 

lanedance

Homework Helper
3,305
2
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...
 

tony873004

Science Advisor
Gold Member
1,748
141
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.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top