What is inverse oflaplace (32)/((s^2+4^2)(s+4)^2) ?

The answer to inverse oflaplace (32)/((s^2+4^2)(s+4)^2) is -1/4 cos(4t)+1/4 e^{-4t}+e^{-4t}t

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inverse oflaplace (32)/((s^2+4^2)(s+4)^2)

inverse laplace

\frac{32}{( s ^{2} + 4 ^{2} ) ( s + 4 ) ^{2}}

- \frac{1}{4} cos (4 t) + \frac{1}{4} e^{- 4 t} + e^{- 4 t} t

Solution steps

L^{- 1} {\frac{32}{( s ^{2} + 4 ^{2} ) ( s + 4 ) ^{2}}}

Take the partial fraction of

\frac{32}{( s ^{2} + 4 ^{2} ) ( s + 4 ) ^{2}} : - \frac{s}{4 ( s ^{2} + 16 )} + \frac{1}{4 ( s + 4 )} + \frac{1}{( s + 4 ) ^{2}}

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

L^{- 1} {\frac{s}{4 ( s ^{2} + 16 )}} : \frac{1}{4} cos (4 t)

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

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

= - \frac{1}{4} cos (4 t) + \frac{1}{4} e^{- 4 t} + e^{- 4 t} t

What is inverse oflaplace (32)/((s^2+4^2)(s+4)^2) ?
The answer to inverse oflaplace (32)/((s^2+4^2)(s+4)^2) is -1/4 cos(4t)+1/4 e^{-4t}+e^{-4t}t