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)

\int x^{5} e^{4 x} d x

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

\frac{d}{d x} (tan (x) + x sec^{2} (x))

sin (tan (x) + sec (x))

\frac{d}{d x} (cos (5 x - \frac{π}{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)