Laplace / inverse laplace transform

[goohu]

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Problem: Find a (limited?) solution to the diff eq.

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At the end of the solution, when you transform $$\frac{-1}{s+1} + \frac{2}{s-3}$$
why doesn't it become $$-e^{-t} + 2e^{3t} $$, t>0 ?

 

[HallsofIvy]

That would be correct if the right hand side of the equation were 0. But the delta functions on the right mean that those are correct only for x greater than certain values so the step functions are needed.

 

[goohu]

What values are those and how are the step functions used?

 

[HallsofIvy]

Do you know what "$$\delta(x)$$" and "$$\delta'(x)$$" mean?