What is inverse oflaplace (3s-5)/((s-1)^2) ?

The answer to inverse oflaplace (3s-5)/((s-1)^2) is 3e^t-e^t*2t

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inverse oflaplace (3s-5)/((s-1)^2)

inverse laplace

\frac{3 s - 5}{( s - 1 ) ^{2}}

3 e^{t} - e^{t} \cdot 2 t

Solution steps

L^{- 1} {\frac{3 s - 5}{( s - 1 ) ^{2}}}

Take the partial fraction of

\frac{3 s - 5}{( s - 1 ) ^{2}} : \frac{3}{s - 1} - \frac{2}{( s - 1 ) ^{2}}

= L^{- 1} {\frac{3}{s - 1} - \frac{2}{( s - 1 ) ^{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{3}{s - 1}} - L^{- 1} {\frac{2}{( s - 1 ) ^{2}}}

L^{- 1} {\frac{3}{s - 1}} : 3 e^{t}

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

= 3 e^{t} - e^{t} \cdot 2 t

f (x) = arctan (2 x)

\int (2 x - 6)^{3} d x

\frac{d}{d x} (\frac{1}{1 + 4 x})

\int_{0}^{3} (x + 1)^{\frac{1}{2}} d x

\frac{\partial}{\partial x} (ln (5 y e^{x y}))

What is inverse oflaplace (3s-5)/((s-1)^2) ?
The answer to inverse oflaplace (3s-5)/((s-1)^2) is 3e^t-e^t*2t