# Inverse laplace transform of this simple function?

what is the inverse laplace transform of (2s)(1/(s-2))?

could i use the identity ∫f(T)g(t-T)dT=F(s)G(s)?

i was hesitant so i figured i'd just ask before i continue..

## Answers and Replies

lurflurf
Homework Helper
write

$$\frac{s}{s-2}=\frac{s-2+2}{s-2}=1+\frac{2}{s-2}$$

or go ahead and use convolution

Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
what is the inverse laplace transform of (2s)(1/(s-2))?

could i use the identity ∫f(T)g(t-T)dT=F(s)G(s)?

i was hesitant so i figured i'd just ask before i continue..

You could use the property
$$\cal{L}(f^\prime)(s) = s\, \cal{L}(f)(s) - f(0+)$$
Be careful, though: that is for a LT defined as
$$\cal{L}(f)(s) = \int_{0+}^{\infty} e^{-st} f(t) \, dt.$$
If, instead, you prefer to define your LT as
$$\cal{L}(f)(s) = \int_{0-}^{\infty} e^{-st} f(t) \, dt,$$
then you will need
$$\cal{L}(f^\prime)(s) = s\, \cal{L}(f)(s) - f(0-)$$