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Laplace / inverse laplace transform

goohu

New member
Dec 29, 2018
19
prob.JPG

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

Sol.JPG



At the end of the solution, when you transform \(\displaystyle \frac{-1}{s+1} + \frac{2}{s-3}\)
why doesnt it become \(\displaystyle -e^{-t} + 2e^{3t} \), t>0 ?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,149
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

New member
Dec 29, 2018
19
What values are those and how are the step functions used?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,149
Do you know what "[tex]\delta(x)[/tex]" and "[tex]\delta'(x)[/tex]" mean?