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

The answer to inverse oflaplace s/(s^4+4) is -1/4 e^{-t}sin(t)+1/4 e^tsin(t)

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inverse oflaplace s/(s^4+4)

inverse laplace

\frac{s}{s ^{4} + 4}

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

Solution steps

L^{- 1} {\frac{s}{s ^{4} + 4}}

Take the partial fraction of

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

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

Expand

\frac{1}{4 ( s ^{2} + 2 s + 2 )} : \frac{1}{4} \cdot \frac{1}{( s + 1 ) ^{2} + 1}

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

Expand

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

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

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)}

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

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

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

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

\int \frac{2 x ^{4} + 4 x ^{3} - x}{x ^{3}} d x

\int 6^{x} e^{x} d x

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{17}{2}}} d x

y^{^{'}} + y = - 4 cos (x)

x \to \infty lim (1^{x})

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