Question

s+3/s²+6s-5=L⁻¹ s+3/(s+3)² -14 when completing the square is used, can u show me the next step to determine the L⁻¹?

Answer 1

\[\large \mathcal L \left\{ \sinh at \right\}=\frac{ s }{ s^2+a^2 }\] and use the first shifting property of the transform \[\large \mathcal L \left\{ e^{at}f(t) \right\}=F(s-a)\]

Answer 2

erm...L{cosh at} or L{sinh at}??

Answer 3

my mistake. it it cosh at, not sinh at.

Answer 4

nvm, btw thx a lot for help k, i really appreciate it.. =)

Answer 5

\[\large F(s+3)= \frac{ s+3 }{ (s+3)^2-14 }\] so \[\large F(s)=\frac{ s }{ s^2-14 }\]