College Math Teaching

November 6, 2013

A weird Laplace Transform (a resonance equation)

Filed under: applied mathematics, calculus, differential equations, Laplace transform — collegemathteaching @ 12:01 am

Ok, we have y" + y = sin(t), y(0) =0, y'(0) = 0 . Now we can solve this by, say, undetermined coefficients and obtain y = \frac{1}{2}sin(t) -\frac{1}{2}tcos(t)

But what happens when we try Laplace Transforms? It is easy to see that the Laplace transform of the equation yields (s^2+1)Y(s)=\frac{1}{s^2+1} which yields Y(s) =\frac{1}{(s^2+1)^2}

So, how do we take the inverse Laplace transform of \frac{1}{(s^2+1)^2}?

Here is one way: we recognize L(tf(t)) = -1\frac{d}{ds}F(s) where L(f(t)) = F(s) .

So, we might try integrating: \int \frac{1}{(s^2+1)^2} ds .

(no cheating with a calculator! 🙂 )

In calculus II, we do: s = tan(\theta), ds = sec^2(\theta) d\theta .

Then \int \frac{1}{(s^2+1)^2} ds is transformed into \int \frac{sec^2(\theta)}{sec^4 \theta} d\theta = \int cos^2(\theta) d \theta = \int \frac{1}{2} + \frac{1}{2}cos(2 \theta) d \theta = \frac{1}{2} \theta + \frac{1}{4}sin(2 \theta) (plus a constant, of course).

We now use sin(2\theta) = 2sin(\theta)cos(\theta) to obtain \frac{1}{2} \theta + \frac{1}{4}sin(2 \theta) = \frac{1}{2} \theta + \frac{1}{2} sin(\theta)cos(\theta) + C .

Fair enough. But now we have to convert back to s . We use tan(\theta) = s to obtain cos(\theta) = \frac{1}{\sqrt{s^2+1}}, sin(\theta) = \frac{s}{\sqrt{s^2+1}}

So \frac{1}{2} \theta + \frac{1}{2} sin(\theta)cos(\theta) converts to \frac{1}{2}arctan(s) + C +\frac{1}{2}\frac{s}{s^2+1} = \int Y(s) ds . Now we use the fact that as s goes to infinity, \int Y(s) has to go to zero; this means C = -\frac{\pi}{2} .

So what is the inverse Laplace transform of \int Y(s) ds ?

Clearly, \frac{1}{2}\frac{s}{s^2+1} gets inverse transformed to \frac{1}{2}cos(t) , so the inverse transform for this part of Y(s) is -\frac{t}{2}cos(t).

But what about the other part? \frac{d}{ds} (arctan(s) - \frac{\pi}{2}) = \frac{1}{1+s^2} so \frac{1}{1+s^2} = -L(tf(t)) which implies that tf(t) = -sin(t) so -tf(t) = sin(t) and so the inverse Laplace transform for this part of Y(s) is \frac{1}{2} sin(t) and the result follows.

Put another way: L(\frac{sin(t)}{t}) =- arctan(s) + C but since we want 0 when s = \infty, C = \frac{\pi}{2} and so L(\frac{sin(t)}{t}) = \frac{\pi}{2}- arctan(s) = arctan(\frac{1}{s}) .

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2 Comments »

  1. […] talked about one way to solve by using Laplace transforms WITHOUT using convolutions; I happen to think that using convolutions is the easiest way […]

    Pingback by Inverse Laplace transform example: 1/(s^2 +b^2)^2 | College Math Teaching — November 6, 2013 @ 11:33 pm

  2. […] we use the Laplace transform, we obtain: which leads to . Now we’ve covered how to do this without convolutions. But the convolution integral is much easier: write which means that […]

    Pingback by Why I teach multiple methods for the inverse Laplace Transform. | College Math Teaching — November 12, 2013 @ 8:33 pm


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