College Math Teaching

November 6, 2013

Inverse Laplace transform example: 1/(s^2 +b^2)^2

Filed under: basic algebra, differential equations, Laplace transform — Tags: — collegemathteaching @ 11:33 pm

I talked about one way to solve y''+y = sin(t), y(0) =y'(0) = 0 by using Laplace transforms WITHOUT using convolutions; I happen to think that using convolutions is the easiest way here.

Here is another non-convolution method: Take the Laplace transform of both sides to get Y(s) = \frac{1}{(s^2+1)^2} .

Now most tables have L(tsin(at)) = \frac{2as}{(s^2 + a^2)^2}, L(tcos(at)) = \frac{s^2-a^2}{(s^2+a^2)^2}

What we have is not in one of these forms. BUT, note the following algebra trick technique:

\frac{1}{s^2+b^2} = (A)(\frac{s^2-b^2}{(s^2 + b^2)^2} - \frac{s^2+b^2}{(s^2+b^2)^2}) when A = -\frac{1}{2b^2} .

Now \frac{s^2-b^2}{(s^2 + b^2)^2} = L(tcos(bt)) and \frac{s^2+b^2}{(s^2+b^2)^2} = \frac{1}{(s^2+b^2)} = L(\frac{1}{b}sin(bt)) and one can proceed from there.


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