# College Math Teaching

## November 12, 2013

### Why I teach multiple methods for the inverse Laplace Transform.

I’ll demonstrate with a couple of examples:

$y''+4y = sin(2t), y(0) = y'(0) = 0$

If we use the Laplace transform, we obtain: $(s^2+4)Y = \frac{2}{s^2+4}$ which leads to $Y = \frac{2}{(s^2+4)^2}$. Now we’ve covered how to do this without convolutions. But the convolution integral is much easier: write $Y = \frac{2}{(s^2+4)^2} = \frac{1}{2} \frac{2}{s^2+4}\frac{2}{s^2+4}$ which means that $y = \frac{1}{2}(sin(2t)*sin(2t)) = \frac{1}{2}\int^t_0 sin(2u)sin(2t-2u)du = -\frac{1}{4}tcos(2t) + \frac{1}{8}sin(2t)$.

Note: if the integral went too fast for you and you don’t want to use a calculator, use $sin(2t-2u) = sin(2t)cos(2u) - cos(2t)sin(2u)$ and the integral becomes $\frac{1}{2}\int^t_0 sin(2t)cos(2u)sin(2u) -cos(2t)sin^2(2u)du =$

$\frac{1}{2} (sin(2t))\frac{1}{4}sin^2(2u)|^t_0 - cos(2t)(\frac{1}{4})( t - \frac{1}{4}sin(4u)|^t_0 =$

$\frac{1}{8}sin^3(2t) - \frac{1}{4}tcos(2t) +\frac{1}{16}sin(4t)cos(2t) =$

$\frac{1}{8}(sin^3(2t) +sin(2t)cos^2(2t))-\frac{1}{4}tcos(2t)$

$= \frac{1}{8}sin(2t)(sin^2(2t) + cos^2(2t))-\frac{1}{4}tcos(2t) = -\frac{1}{4}tcos(2t) + \frac{1}{8}sin(2t)$

Now if we had instead: $y''+4y = sin(t), y(0)=0, y'(0) = 0$

The Laplace transform of the equation becomes $(s^2+4)Y = \frac{1}{s^2+1}$ and hence $Y = \frac{1}{(s^2+1)(s^2+4)}$. One could use the convolution method but partial fractions works easily: one can use the calculator (“algebra” plus “expand”) or:

$\frac{A+Bs}{s^2+4} + \frac{C + Ds}{s^2+1} =\frac{1}{(s^2+4)(s^2+1)}$. Get a common denominator and match numerators:

$(A+Bs)(s^2+1) + (C+Ds)(s^2+4) = 1$. One can use several methods to resolve this: here we will use $s = i$ to see $(C + Di)(3) = 1$ which means that $D = 0$ and $C = \frac{1}{3}$. Now use $s = 2i$ so obtain $(A + 2iB)(-3) = 1$ which means that $B = 0, A = -\frac{1}{3}$ so $Y = \frac{1}{3} (\frac{1}{s^2+1} - \frac{1}{s^2+4}$ so $y = \frac{1}{3} (sin(t) - \frac{1}{2} sin(2t)) = \frac{1}{3}sin(t) -\frac{1}{6}sin(2t)$

So, sometimes the convolution leads us to the answer quicker than other techniques and sometimes other techniques are easier.