Answer to Question #233089 in Math for abi

Question #233089

L-1{ s+1 / (s2+2s+s)2 }


1
Expert's answer
2021-09-07T18:43:15-0400
"\\dfrac{s+1}{(s^2+2s+s)^2}=\\dfrac{s+1}{(s^2+3s)^2}"

"=\\dfrac{A}{s}+\\dfrac{B}{s^2}+\\dfrac{C}{s+3}+\\dfrac{D}{(s+3)^2}"

"=\\dfrac{As(s+3)^2+B(s+3)^2+Cs^2(s+3)+Ds^2}{(s^2+3s)^2}"

"=\\dfrac{As^3+6As^2+9As+Bs^2+6Bs+9B+Cs^3+3Cs^2+Ds^2}{(s^2+3s)^2}"

"s^3:A+C=0"


"s^2:6A+B+3C+D=0"


"s^1:9A+6B=1"


"s^0:9B=1"



"A=\\dfrac{1}{27}, B=\\dfrac{1}{9}, C=-\\dfrac{1}{27}, D=-\\dfrac{2}{9}""L^{-1}(\\dfrac{s+1}{(s^2+2s+s)^2})=\\dfrac{1}{27}L^{-1}(\\dfrac{1}{s})+\\dfrac{1}{9}L^{-1}(\\dfrac{1}{s^2})"

"-\\dfrac{1}{27}L^{-1}(\\dfrac{1}{s+3})-\\dfrac{2}{9}L^{-1}(\\dfrac{1}{(s+3)^2})"

"=\\dfrac{1}{27}+\\dfrac{t}{9}-\\dfrac{1}{27}e^{-3t}-\\dfrac{2}{9}te^{-3t}"




"L^{-1}(\\dfrac{s+1}{(s^2+2s+s)^2})=\\dfrac{1}{27}+\\dfrac{t}{9}-\\dfrac{1}{27}e^{-3t}-\\dfrac{2}{9}te^{-3t}"




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