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I'm working on solving linear ordinary differential equations with Laplace transforms, specifically where the equations involves the Heaviside function. The equation I'm working on is

y'' + y = u(t-3),

where u(t-3) is the Heaviside function shifted so the discontinuity is at t=3. I get the answer, but I'm off by a factor of the Heaviside function. What I get is

y = 1 - cos (t-3) + sin t, but the answer in the book is

y = u(t-3)[1 - cos (t-3)] + sin t.

So I take the Laplace transform of both sides and solve for Y(s); I get

Y(s) = (s^2 + 1)^-1(s^-1)e^(-3s) + (s^2 + 1)^-1.

The inverse Laplace of the last term is sin t. To get the inverse Laplace of the first term, I use the convolution theorem and the fact that u(t-3)=0 for t<3 and u(t-3)=1 for t>3. I get this

ʃu(t-3)sin(t-v)dv from 0 to t

= ʃsin(t-v)dv from 3 to t

= cos (t-v) evaluated from 3 to t

= 1 - cos (t-3).

I think I'm doing something incorrect in the integral, but I can't figure it out. Does the convolution theorem somehow not apply to the Heaviside function? The book I'm using has absolutely no examples involving the convolution of the Heaviside function and from the notes I took when I had the class, we didn't go over any in there as well.

You're right -- you're doing the integral wrong.

Now you are right, if t>3 then it makes perfect sense for you to write ʃsin(t-v)dv = cos(t-v)

The problem is that this computation is false if t<3.

If t<3, then the integral is 0 because you're only integrating up to some value of t before the integrand ever becomes nonzero. For this reason, what you have is:

ʃu(t-3)sin(t-v)dv from 0 to t =

0 if t<3

1-cos(t-3) if t>=3

But what is that? It is, literally, the same as u(t-3)(1-cos(t-3)). That's what a factor of u(t-3) will do for you.

What you really need to do is consider that u(t-3) is a piecewise-defined function, and consider the integration piecewise as well. You've done that correctly -- but it's important to make sure that you get 0 when t<3.

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