Laplace Transformation with a sum of a Dirac Delta Function

Question

Find the solution of the following IVP:

1. y''(t)+y(t)= ∑(k=1 to 10) (-1)^(k+1)*(−)
2. y(0)
3. y'(0)=0

So using the Laplace Transformation:

s^2Y(s)+Y(s)= ∑(k=1 to 10)(-1)^(k+1)* exp(-k*pi *s)/(s^2+1)

And isolating and stuff I would have:

y(t)= ∑(k=1 to 10)(-1)^(k+1)* Heaviside[kpi] sin(t-k*pi)

Basiaclly what I want to know is if I can even do it like this, or do I have to work with a diffrent approach? Sorry for the horrible formating I have no idea how to do it better tbh. I appreciate any help I can get!

Answer 1

We are given

$$y''(t)+y(t)= \sum_{k=1}^{10} (-1)^{k+1}\delta(−\pi k)\\y(0)=0\\y'(0)=0$$

First, recall that the Laplace operator is linear. Therefore,

$$\displaystyle\mathcal{L}\left(\sum_{k=1}^{10} (-1)^{k+1}\delta(−\pi k)\right) = \sum_{k=1}^{10} \mathcal{L}\left((-1)^{k+1}\delta(−\pi k)\right) = \sum_{k=1}^{10} \left((-1)^{k+1} e^{-k \pi s}\right) $$

Next

$$\mathcal{L}(y'' + y) = s^2 Y(s) - s Y(0) - Y'(0) + Y(s) = (s^2+1) Y(s)$$

Putting it all together

$$ (s^2+1) Y(s) = \sum_{k=1}^{10} \left((-1)^{k+1} e^{-k \pi s}\right)$$

So

$$Y(s) = \sum_{k=1}^{10}\left(\dfrac{(-1)^{k+1} e^{-k \pi s}}{s^2+1}\right)$$

Please continue with the inverse Laplace transform.

Spoiler (from the previous step, you can simplify things, but your approach is okay)

$$y(t) = -\sum_{k=1}^{10}\sin(t) u(t - k \pi)$$

Answer 2

Consider the first impulse δ(t- π) whose Laplace transform (LT)is exp(-sπ). Taking the LT of 2nd order DE, we have s^2 Y(s)- s y'(0-) -y(0-) + Y(s) = exp (- s π). Collecting the like terms and noting y'(0)=0 and y(0-)=y0 we have (s^2+1) Y(s) = y0 + exp(-s π). Hence Y(s)= y0/(s^2+1) + exp(-s π)/(s^2+1). Taking inverse transform, we have y(t) = y0 * sin t * u(t) + sin (t-π) u(t-π))

Since we have a LTI system, adding more impulses -δ(t-2 π)+δ(t- 3π)-... -δ(t- 10π) simply means we are adding more corresponding impulse response. Thus we have y(t) = y0 * sin t * u(t) + sin (t-π) u(t-π) -sin (t-2π) u(t-2π) +... -sin(t-10π) * u(t-10π) = y0 *sin t *u(t) + ∑(k=1 to 10) { (-1)^(k+1) * sin (t-kπ) u(t-kπ)} = y0 *sin t u(t) + ∑(k=1 to 9) { (-1)^(k+1) * k * sin (t-kπ) [u(t-kπ) - u(t-(k+1) π)]} - 10 *sin(t-10π) u(t-10π)