Question #1dfad
2 Answers
Explanation:
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Due to
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# ℒ^(-1) \ { 1/sqrt(2s+3) } = e^(-3/2t) /sqrt(2pit) #
Explanation:
We seek:
# ℒ^(-1) \ { 1/sqrt(2s+3) } #
We will need the following standard Laplace transform and inverses:
# {: (ul(f(t)=ℒ^(-1){F(s)}), ul(F(s)=ℒ{f(t)}), ul("Notes")), (af(t), aF(s),a " constant"), (e^(at){1/sqrt(pit)-be^(b^2t)erfc(bsqrt(t))}, 1/(sqrt(s-a)+b),a","b " constant") :} #
So we can write:
# ℒ^(-1) \ { 1/sqrt(2s+3) } = ℒ^(-1) \ { 1/sqrt(2(s+3/2)) } #
# " " = 1/sqrt(2) \ ℒ^(-1) \ { 1/sqrt(s+3/2) } #
# " " = 1/sqrt(2) \ e^(-3/2t){1/sqrt(pit)-0} \ \ \ (a=-3/2,b=0)#
# " " = e^(-3/2t) /sqrt(2pit) #