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

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

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

inverse laplace

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

δ (t) - 4 e^{- 2 t}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {1} : δ (t)

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

= δ (t) - 4 e^{- 2 t}

\frac{\partial}{\partial y} (2 sec^{2} (2 x + 3 y))

\int 4 - 9 x^{2} d x

e^{x - y}

y = 7

\int x^{2} (x^{3} + 1)^{\frac{1}{2}} d x

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