1. ## Delta Function

Find, graph, and discuss the solution of the following:

y'' + y = delta(t - 2pi)
y'(0) = 0
y(0) = 10

where delta is Dirac's delta function

Transforming, we get:

(s^2)Y + Y = e^(-2*pi*s)
(s^2 + 1)Y = e^(-2*pi*s)
Y = [e^(-2*pi*s)]/[(s^2 + 1)]
y(t) = L^-1[Y(s)]

Assuming everything's correct up to this point, I can see sin(t) in the denominator of Y above, but the answer in the back of the book is:
y = 10cos(t) if 0 < t < 2*pi
y = 10cos(t) + sin(t) if t > 2*pi

I really feel uncomfortable with my knowledge of time-shifts, so if someone could please help explain how to get from point A to point B, it'd be much appreciated. Not looking for complete solution, but enough to get this horse to see where the water might be located. :-)

2. Originally Posted by dsprice
Find, graph, and discuss the solution of the following:

y'' + y = delta(t - 2pi)
y'(0) = 0
y(0) = 10

where delta is Dirac's delta function

Transforming, we get:

(s^2)Y + Y = e^(-2*pi*s)
(s^2 + 1)Y = e^(-2*pi*s)
Y = [e^(-2*pi*s)]/[(s^2 + 1)]
y(t) = L^-1[Y(s)]

Assuming everything's correct up to this point, I can see sin(t) in the denominator of Y above, but the answer in the back of the book is:
y = 10cos(t) if 0 < t < 2*pi
y = 10cos(t) + sin(t) if t > 2*pi

I really feel uncomfortable with my knowledge of time-shifts, so if someone could please help explain how to get from point A to point B, it'd be much appreciated. Not looking for complete solution, but enough to get this horse to see where the water might be located. :-)
The laplace transform is incorrect

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

So the transformed equation should be

$\displaystyle s^2-10s+Y=e^{2\pi s}$ solving for Y gives

$\displaystyle Y=10\frac{s}{s^2+1}+\frac{e^{2\pi s}}{s^2+1}$ Inverting the transform gives

$\displaystyle y(t)=10\cos(t)+\mathcal{U}(t-2\pi)\sin(t-2\pi)$

Where $\displaystyle \mathcal{U}(t)$ is the unit step function.

Note this is the same as

$\displaystyle y(t) = \begin{cases}10\cos(t), \text{ if }0 \le t \le 2 \pi \\ 10\cos(t)+\sin(t), \text{ if } t > 2\pi \end{cases}$

3. Thanks! The missing Y in:

threw me off for a while, but that's actually good because in working through it, it's starting to make sense again. I'm sure it did 25 years ago during my undergrad...so slowly but surely the cobwebs are clearing. :-)