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

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

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

inverse laplace

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

\frac{e ^{- 2 t} t}{2} + \frac{e ^{2 t} t}{2}

Solution steps

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

Take the partial fraction of

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

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

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

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

= \frac{e ^{- 2 t} t}{2} + \frac{e ^{2 t} t}{2}

5 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0

2 ln (\frac{3 x ^{2}}{2})

\frac{d}{d x} ((4 x^{2} + 9) (8 x + x^{\frac{1}{2}}))

y = 7 x^{\frac{1}{2}} + x^{\frac{3}{2}}

\int \frac{5 x}{3 x ^{2} + 4} d x

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