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

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

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inverse oflaplace-(e^{(-2s)})/((s^2-4s+4))

inverse laplace

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

- H (t - 2) e^{2 (t - 2)} (t - 2)

Solution steps

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

Factor

s^{2} - 4 s + 4 : (s - 2)^{2}

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

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

= - H (t - 2) e^{2 (t - 2)} (t - 2)

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