What is inverse oflaplace 3/((s+4)(s+8)^2) ?

The answer to inverse oflaplace 3/((s+4)(s+8)^2) is 3/16 e^{-4t}-3/16 e^{-8t}-(3e^{-8t}t)/4

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inverse oflaplace 3/((s+4)(s+8)^2)

inverse laplace

\frac{3}{( s + 4 ) ( s + 8 ) ^{2}}

\frac{3}{16} e^{- 4 t} - \frac{3}{16} e^{- 8 t} - \frac{3 e ^{- 8 t} t}{4}

Solution steps

L^{- 1} {\frac{3}{( s + 4 ) ( s + 8 ) ^{2}}}

Take the partial fraction of

\frac{3}{( s + 4 ) ( s + 8 ) ^{2}} : \frac{3}{16 ( s + 4 )} - \frac{3}{16 ( s + 8 )} - \frac{3}{4 ( s + 8 ) ^{2}}

= L^{- 1} {\frac{3}{16 ( s + 4 )} - \frac{3}{16 ( s + 8 )} - \frac{3}{4 ( s + 8 ) ^{2}}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{3}{16 ( s + 4 )}} - L^{- 1} {\frac{3}{16 ( s + 8 )}} - L^{- 1} {\frac{3}{4 ( s + 8 ) ^{2}}}

L^{- 1} {\frac{3}{16 ( s + 4 )}} : \frac{3}{16} e^{- 4 t}

L^{- 1} {\frac{3}{16 ( s + 8 )}} : \frac{3}{16} e^{- 8 t}

L^{- 1} {\frac{3}{4 ( s + 8 ) ^{2}}} : \frac{3 e ^{- 8 t} t}{4}

= \frac{3}{16} e^{- 4 t} - \frac{3}{16} e^{- 8 t} - \frac{3 e ^{- 8 t} t}{4}

\int (cos (5 x)) d x

\frac{d}{d x} (2^{l n (x)})

y = cos (a^{4} + x^{4})

(7, 5), (- 4, - 1)

\int_{0}^{\frac{π}{4}} sec^{5} (x) d x

What is inverse oflaplace 3/((s+4)(s+8)^2) ?
The answer to inverse oflaplace 3/((s+4)(s+8)^2) is 3/16 e^{-4t}-3/16 e^{-8t}-(3e^{-8t}t)/4