Question

at http://tutorial.math.lamar.edu/Classes/DE/StepFunctions.aspx after 1st example while apply Laplace transform on Heaviside function the author has put f(t) =1in equation (2) whereas that equation has f(t-c) how could he put f(t)'s value of "1" in f(t-c) as "1", please someone help me in understanding it :|

Answer 1

I don't think I understand your question...

Answer 2

is it because we can't transofrm f(t-c) therefore we are applying L{1} on e^-cs?

Answer 3

oh yes, now I get your question...
yes your answer is basically the correct one

Answer 4

@TuringTest please take a look at the attached pic

Answer 5

yeah, exactly
the laplace of the heaviside function is known only int the form \[\mathcal L\{u_c(t)f(t-c)\}=e^{-cs}F(s)\]so we need to recognize the function multiplying the heaviside in order to take its laplace. In this case f(t-c)=1

Answer 6

(to last Q) yes, what about it?

Answer 7

function is known only int the form ???? "int" you mean integral?

Answer 8

no, typo
*known in the form...

Answer 9

okkkkkkkkkkk

Answer 10

@turing thankssssssssss a tonne bro

Answer 11

anytime :D