What is inverse oflaplace 1/((s^4+1)) ?

The answer to inverse oflaplace 1/((s^4+1)) is

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inverse oflaplace 1/((s^4+1))

inverse laplace

\frac{1}{( s ^{4} + 1 )}

\frac{1}{2 2} e^{- \frac{t}{2}} cos (\frac{1}{2} t) + \frac{1}{2 2} e^{- \frac{t}{2}} sin (\frac{1}{2} t) - \frac{1}{2 2} e^{\frac{t}{2}} cos (\frac{1}{2} t) + \frac{1}{2 2} e^{\frac{t}{2}} sin (\frac{1}{2} t)

Solution steps

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

Take the partial fraction of

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

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

Expand

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

= L^{- 1} ⎩ ⎨ ⎧ \frac{1}{2 2} \cdot \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{1}{2}} + \frac{1}{4} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{2}} + \frac{- s + 2}{2 2 ( s ^{2} - 2 s + 1 )} ⎭ ⎬ ⎫

Expand

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

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

= \frac{1}{2 2} L^{- 1} ⎩ ⎨ ⎧ \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{1}{2}} ⎭ ⎬ ⎫ + \frac{1}{4} L^{- 1} ⎩ ⎨ ⎧ \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{2}} ⎭ ⎬ ⎫ - \frac{1}{2 2} L^{- 1} ⎩ ⎨ ⎧ \frac{s - \frac{1}{2}}{( s - \frac{1}{2} ) ^{2} + \frac{1}{2}} ⎭ ⎬ ⎫ + \frac{1}{4} L^{- 1} ⎩ ⎨ ⎧ \frac{1}{( s - \frac{1}{2} ) ^{2} + \frac{1}{2}} ⎭ ⎬ ⎫

L^{- 1} ⎩ ⎨ ⎧ \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{1}{2}} ⎭ ⎬ ⎫ : e^{- \frac{t}{2}} cos (\frac{1}{2} t)

L^{- 1} ⎩ ⎨ ⎧ \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{2}} ⎭ ⎬ ⎫ : e^{- \frac{t}{2}} 2 sin (\frac{1}{2} t)

L^{- 1} ⎩ ⎨ ⎧ \frac{s - \frac{1}{2}}{( s - \frac{1}{2} ) ^{2} + \frac{1}{2}} ⎭ ⎬ ⎫ : e^{\frac{t}{2}} cos (\frac{1}{2} t)

L^{- 1} ⎩ ⎨ ⎧ \frac{1}{( s - \frac{1}{2} ) ^{2} + \frac{1}{2}} ⎭ ⎬ ⎫ : e^{\frac{t}{2}} 2 sin (\frac{1}{2} t)

= \frac{1}{2 2} e^{- \frac{t}{2}} cos (\frac{1}{2} t) + \frac{1}{4} e^{- \frac{t}{2}} 2 sin (\frac{1}{2} t) - \frac{1}{2 2} e^{\frac{t}{2}} cos (\frac{1}{2} t) + \frac{1}{4} e^{\frac{t}{2}} 2 sin (\frac{1}{2} t)

Refine

\frac{1}{2 2} e^{- \frac{t}{2}} cos (\frac{1}{2} t) + \frac{1}{4} e^{- \frac{t}{2}} 2 sin (\frac{1}{2} t) - \frac{1}{2 2} e^{\frac{t}{2}} cos (\frac{1}{2} t) + \frac{1}{4} e^{\frac{t}{2}} 2 sin (\frac{1}{2} t) : \frac{1}{2 2} e^{- \frac{t}{2}} cos (\frac{1}{2} t) + \frac{1}{2 2} e^{- \frac{t}{2}} sin (\frac{1}{2} t) - \frac{1}{2 2} e^{\frac{t}{2}} cos (\frac{1}{2} t) + \frac{1}{2 2} e^{\frac{t}{2}} sin (\frac{1}{2} t)

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

y = 2 (x + 1), y = 3 (x + 1), x = 6

6 e^{x} + x

\frac{s}{s ^{2} - π}

y = x^{12} e^{x}

f (x) = \frac{cos ( x )}{x ^{6}}

What is inverse oflaplace 1/((s^4+1)) ?
The answer to inverse oflaplace 1/((s^4+1)) is