# 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 $L(\frac{sin(t)}{t}) = \frac{\pi}{2}- arctan(s) = arctan(\frac{1}{s})$ .

## 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