# Graphing a Heaviside unit function

[SOLVED] Graphing a Heaviside unit function

1. Homework Statement
find solution of the following differential eq and graph it:

$$y'' + 4y = \delta(t-2\pi)$$
$$y(0)=0$$
$$y'(0)=0$$

2. Homework Equations
$$\delta$$
is the Dirac delta function
$$u_{c}$$
is the Heaviside unit step function

3. The Attempt at a Solution

I used the laplace transform and found the solution to be:

$$\frac{1}{2}u_{\pi}(sin(2(t-\pi)))-\frac{1}{2}u_{2\pi}(sin(2(t-2\pi)))$$

which i checked and it is right. However, I'm not sure how to graph this. The following is what I have:

before t = pi and after t = 2pi, y = 0.
But, in between what will it be? If anyone could please help me out I would greatly appreciate it.

Related Calculus and Beyond Homework Help News on Phys.org
As the sine function is periodic, you can write the expression as
$$\frac{1}{2}sin2t(u_{\pi}-u_{2\pi})$$
Now can you proceed?

Thank you for the help! So, then from pi to 2 pi I will just have that sine wave? Also, if you could please help me figure this out:

so then is this:
$$\frac{1}{2}u_{\pi}(sin(2(t-\pi)))-\frac{1}{2}u_{2\pi}(sin(2(t-2\pi)))$$
not the same as this?:
$$\frac{1}{2}u_{\pi}(sin(2(t-\pi)))+\frac{1}{2}u_{2\pi}(-sin(2(t-2\pi)))$$

Thank you for the help! So, then from pi to 2 pi I will just have that sine wave?
Correct :)

About the second question, yes you can take the minus sign with the sine term, that's basic multiplication. sin(2t-2pi) = sin(2t-4pi) = sin(2t)

Thanks again, but i'm not sure that i stated the second question right. I mean that if i bring the minus sign inside of the parenthesis it implies to me (maybe incorrectly) that the second sine wave will start at 2pi and propagate to infinity.

Okay let me clear things up a bit,
$$\frac{1}{2}u_{\pi}(sin(2(t-\pi)))-\frac{1}{2}u_{2\pi}(sin(2(t-2\pi)))$$
$$\frac{1}{2}sin(2t)u_{\pi}-\frac{1}{2}sin(2t)u_{2\pi}$$