# Laplace / inverse laplace transform

#### goohu

##### New member

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

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

#### HallsofIvy

##### Well-known member
MHB Math Helper
Do you know what "$$\delta(x)$$" and "$$\delta'(x)$$" mean?