What is inverse oflaplace (10-4s)/((s-2)^2) ?

The answer to inverse oflaplace (10-4s)/((s-2)^2) is -4e^{2t}+e^{2t}*2t

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inverse oflaplace (10-4s)/((s-2)^2)

inverse laplace

\frac{10 - 4 s}{( s - 2 ) ^{2}}

- 4 e^{2 t} + e^{2 t} \cdot 2 t

Solution steps

L^{- 1} {\frac{10 - 4 s}{( s - 2 ) ^{2}}}

Take the partial fraction of

\frac{10 - 4 s}{( s - 2 ) ^{2}} : - \frac{4}{s - 2} + \frac{2}{( s - 2 ) ^{2}}

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

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

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

= - 4 e^{2 t} + e^{2 t} \cdot 2 t

\frac{\partial}{\partial x} (- 2 x e^{- 2 x y})

\int_{0}^{π} sec^{2} (\frac{t}{4}) d t

\int x arcsin (x) d x

\frac{d}{d x} ((cos (x))^{- 1})

\int (- 14 sin (x)) d x

What is inverse oflaplace (10-4s)/((s-2)^2) ?
The answer to inverse oflaplace (10-4s)/((s-2)^2) is -4e^{2t}+e^{2t}*2t