Laplace Transformation with a sum of a Dirac Delta Function Find the solution of the following IVP:

*

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

*y(0)

*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!
 A: 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)

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

A: 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)
