Answer to Question #343126 in Differential Equations for Tobias Felix

Question #343126

Find the Laplace transforms of the following function:

8. L{t^2 e^-2t + e^-t cos2t + 3}


1
Expert's answer
2022-05-24T17:58:26-0400
"F(s)=L(t^2e^{-2t}+e^{-t}\\cos(2t)+3)"

"=\\displaystyle\\int_{0}^{\\infin}e^{-st}(t^2e^{-2t}+e^{-t}\\cos(2t)+3)dt"

"\\int t^2e^{-(s+2)t}dt=-\\dfrac{t^2e^{-(s+2)t}}{s+2}+\\dfrac{2}{s+2}\\int te^{-(s+2)t}dt"

"=-\\dfrac{t^2e^{-(s+2)t}}{s+2}-\\dfrac{2te^{-(s+2)t}}{(s+2)^2}"


"+\\dfrac{2}{(s+2)^2}\\int e^{-(s+2)t}dt"

"=-\\dfrac{t^2e^{-(s+2)t}}{s+2}-\\dfrac{2te^{-(s+2)t}}{(s+2)^2}-\\dfrac{2e^{-(s+2)t}}{(s+2)^3}+C"


"L(t^2e^{-2t})=\\lim\\limits_{A\\to\\infin}\\displaystyle\\int_{0}^{\\infin}e^{-st}(t^2e^{-2t})dt"

"=[-\\dfrac{t^2e^{-(s+2)t}}{s+2}-\\dfrac{2te^{-(s+2)t}}{(s+2)^2}-\\dfrac{2e^{-(s+2)t}}{(s+2)^3}]\\begin{matrix}\n A \\\\\n 0\n\\end{matrix}"

"=-0-0-0-(-0-0-\\dfrac{2}{(s+2)^3})=\\dfrac{2}{(s+2)^3}"

"\\int e^{-(s+1)t}\\cos(2t)dt=-\\dfrac{e^{-(s+1)t}\\cos(2t)}{s+1}"

"-\\dfrac{2}{s+1}\\int e^{-(s+1)t}\\sin (2t)dt=-\\dfrac{e^{-(s+1)t}\\cos(2t)}{s+1}"

"+\\dfrac{2e^{-(s+1)t}\\sin(2t)}{(s+1)^2}-\\dfrac{4}{(s+1)^2}\\int e^{-(s+1)t}\\cos (2t)dt"

"((s+1)^2+4)\\int e^{-(s+1)t}\\cos (2t)dt"

"=-e^{-(s+1)t}\\cos(2t)(s+1)+2e^{-(s+1)t}\\sin(2t)"

"\\int e^{-(s+1)t}\\cos (2t)dt=-\\dfrac{e^{-(s+1)t}\\cos(2t)(s+1)}{s^2+2s+5}"

"+\\dfrac{2e^{-(s+1)t}\\sin(2t)}{s^2+s+5}+C_1"


"L(e^{-t}\\cos(2t))=\\lim\\limits_{A\\to\\infin}\\displaystyle\\int_{0}^{\\infin}e^{-st}(e^{-t}\\cos(2t))dt"

"=[-\\dfrac{e^{-(s+1)t}\\cos(2t)(s+1)}{s^2+2s+5}+\\dfrac{2e^{-(s+1)t}\\sin(2t)}{s^2+s+5}]\\begin{matrix}\n A \\\\\n 0\n\\end{matrix}"

"=-0+0-(-\\dfrac{s+1}{s^2+2s+5}+0)=\\dfrac{s+1}{s^2+2s+5}"



"L(3)=\\lim\\limits_{A\\to\\infin}\\displaystyle\\int_{0}^{\\infin}e^{-st}(3)dt"

"=[-\\dfrac{3e^{-st}}{s}+]\\begin{matrix}\n A \\\\\n 0\n\\end{matrix}=-0+\\dfrac{3}{s}=\\dfrac{3}{s}"


"L(t^2e^{-2t}+e^{-t}\\cos(2t)+3)"

"=\\dfrac{2}{(s+2)^3}+\\dfrac{s+1}{s^2+2s+5}+\\dfrac{3}{s}"


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