find-Laplace-transform-t-3-cos-4t-

Question Number 88169 by jagoll last updated on 08/Apr/20

find Laplace transform

t^{3} . \cos 4 t

Commented by mathmax by abdo last updated on 08/Apr/20

L (x^{3} c o s (4 x)) = \int_{0}^{\infty} f (t) e^{- x t} d t

= \int_{0}^{\infty} t^{3} c o s (4 t) e^{- x t} d t = R e (\int_{0}^{\infty} t^{3} e^{i 4 t - x t} d t)

\int_{0}^{\infty} t^{3} e^{(- x + 4 i) t} d t =_{b y p s r t s} {[\frac{t^{3}}{- x + 4 i} e^{(- x + 4 i) t}]}_{0}^{\infty} - \frac{1}{- x + 4 i} \int_{0}^{\infty} 3 t^{2} e^{(- x + 4 i) t} d t

= \frac{3}{x - 4 i} {{[\frac{t^{2}}{- x + 4 i} e^{(- x + 4 i) t}]}_{0}^{\infty} - \frac{1}{- x + 4 i} \int_{0}^{\infty} 2 t e^{(- x + 4 i) t} d t}

= \frac{6}{{(x - 4 i)}^{2}} \int_{0}^{\infty} t e^{(- x + 4 i) t} d t

= \frac{6}{{(x - 4 i)}^{2}} {{[- \frac{t}{- x + 4 i} e^{(- x + 4 i) t}]}_{0}^{\infty} - \frac{1}{- x + 4 i} \int_{0}^{\infty} e^{(- x + 4 i) t} d t}

= \frac{6}{{(x - 4 i)}^{3}} {[\frac{1}{- x + 4 i} e^{(- x + 4 i) t}]}_{0}^{\infty} = \frac{6}{{(x - 4 i)}^{4}} = 6 {(\frac{1}{x - 4 i})}^{4}

= 6 {(\frac{x + 4 i}{x^{2} + 16})}^{4} = \frac{6}{x^{2} + 16} {(x + 4 i)}^{2} {(x + 4 i)}^{2}

= \frac{6}{x^{2} + 16} {(x^{2} + 8 i x - 16)}^{2}

= \frac{6}{x^{2} + 16} {{(x^{2} + 8 i x)}^{2} - 32 (x^{2} + 8 i x) + 16^{2}}

= \frac{6}{x^{2} + 16} {x^{4} + 16 {i x}^{3} - 64 x^{2} - 32 x^{2} - 8.32 i x + 16^{2}} \Rightarrow

L (x^{3} (4 x)) = \frac{6}{x^{2} + 16} (x^{4} - 96 x^{2} + 16^{2})

Commented by mathmax by abdo last updated on 09/Apr/20

f o r g i v e L (x^{3} c o s (4 x)) = \frac{6 (x^{4} - 96 x^{2} + 16^{2})}{{(x^{2} + 16)}^{4}}

Answered by jagoll last updated on 09/Apr/20

L (\cos 4 t) = \frac{s}{s^{2} + 16}

L (t \cos 4 t) = - \frac{d}{ds} [\frac{s}{s^{2} + 16}] = \frac{16 - s^{2}}{{(s^{2} + 16)}^{2}}

L (t^{2} \cos 4 t) = - \frac{d}{ds} [\frac{16 - s^{2}}{{(s^{2} + 16)}^{2}}]

= \frac{96 s - {2 s}^{3}}{{(s^{2} + 16)}^{3}}

L (t^{3} \cos 4 t) = - \frac{d}{ds} [\frac{96 s - {2 s}^{3}}{{(s^{2} + 16)}^{3}}]

= \frac{6 (s^{4} - {96 s}^{2} + 256)}{{(s^{2} + 16)}^{4}}

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