Solving Laplace Transform: Finding \frac{1}{{\left( {s + 4} \right)^2 }}

[tony873004]

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

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

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

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]

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:
$$L(e^{-\alpha t}) = \frac{1}{{\left( {s + 4} \right)^2 }}$$
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]

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.