Answer to Question #343135 in Differential Equations for Tobias Felix

Question #343135

Inverse Laplace Transforms

Find L^-1 {F(s)} when F(s) is given by


5. 1/(s+1)(s+2)(s^2+2s+10)


1
Expert's answer
2022-05-29T16:07:55-0400

5.


"\\dfrac{1}{(s+1)(s+2)(s^2+2s+10)}=\\dfrac{A}{s+1}+\\dfrac{B}{s+2}"




"+\\dfrac{Cs+D}{s^2+2s+10}=\\dfrac{A(s+2)(s^2+2s+10)}{(s+1)(s+2)(s^2+2s+10)}"


"+\\dfrac{B(s+1)(s^2+2s+10)}{(s+1)(s+2)(s^2+2s+10)}"

"+\\dfrac{(Cs+D)(s+1)(s+2)}{(s+1)(s+2)(s^2+2s+10)}"

"s=-1:A(1-2+10)=1"

"s=-2:-B(4-4+10)=1"

"s=0: 20A+10B+2D=1"


"s=1:39A+26B+6C+6D=1"

"A=\\dfrac{1}{9}, B=-\\dfrac{1}{10}, D=-\\dfrac{1}{9}, C=-\\dfrac{1}{90}"


"L^{-1}(\\dfrac{1}{(s+1)(s+2)(s^2+2s+10)})"

"=\\dfrac{1}{9}L^{-1}(\\dfrac{1}{s+1})-\\dfrac{1}{10}L^{-1}(\\dfrac{1}{s+2})"

"-\\dfrac{1}{90}L^{-1}(\\dfrac{s+1}{(s+1)^2+3^2})-\\dfrac{1}{10}L^{-1}(\\dfrac{1}{(s+1)^2+3^2})"



"=\\dfrac{1}{9}e^{-t}-\\dfrac{1}{10}e^{-2t}-\\dfrac{1}{90}e^{-t}\\cos(3t)-\\dfrac{1}{30}e^{-t}\\sin (3t)"




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