Laplace transform Heaviside step function

Pietair
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Homework Statement


What is the Laplace transform of f(t) = 0 for 0 < t < 2 and f(t) = (4-t) for 2 < t < 3 and f(t) = 1 for 3 < t < 4 and f(t) = (5-t) for 4 < t < 5 and f(t) = 0 for t > 5?

Homework Equations


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The Attempt at a Solution


f(t) = H(t-2)(4-t) - H(t-3)(4-t) + H(t-3) - H(t-4) + H(t-4)(5-t) - H(t-5)(5-t)
f(t) = H(t-2)(4-t) + H(t-3)(2t-7) - H(t-4)(t-4) + H(t-5)(t-5)

How can I rewrite H(t-2)(4-t) + H(t-3)(2t-7) to a form suitable for Laplace transform?
 
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For H(t-2)(4-t), you'd like to see a t-2 in the second factor right? So just put one there, and balance it out elsewhere in the expression. Like this:

H(t-2)(4-(t-2)-2)

See how I added and subtracted 2? This simplifies to:

H(t-2)(2-(t-2))

Do the same for H(t-3)(2t-7).
 
Thank you for your reply!

Now I am getting:
<br /> H(t-2)(2-(t-2)) + H(t-3)(2(t-3)-1) - H(t-4)(t-4) + H(t-5)(t-5)<br />

And the Laplace transform:
<br /> ((2/s)-(1/s^2))e^(-2s) + ((2/s^2)-(1/s))e^(-3s) + ((-1/s^2)e^(-4s)) + ((1/s^2)e^(-5s))<br />

Is this correct?
 
Looks good to me.
 
Thanks a lot!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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