# 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.