Laplace Transform


Science Advisor
Gold Member
From the class notes:
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 }} \\

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.


Homework Helper
Hi Tony

The laplace transform of a product is not just the Laplace transform of the components, have a look at this table:
[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
these can be derived using integration by parts on successive derivatives if i remember rightly...


Science Advisor
Gold Member
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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