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

The answer to inverse oflaplace 1/(s-1(s^2)) is \H(t)-e^t

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

inverse laplace

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

H (t) - e^{t}

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {\frac{1}{s} - \frac{1}{s - 1}}

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{1}{s}} - L^{- 1} {\frac{1}{s - 1}}

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

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

= H (t) - e^{t}

\int e^{2 l n (x)} d x

\int (8 x + 5) d x

\int \frac{x}{( 5 - x ^{2} ) ^{2}} d x

36 y^{^{''}} + 60 y^{^{'}} + 25 y = 0

\int_{1}^{e} ln (7 x) d x

What is inverse oflaplace 1/(s-1(s^2)) ?
The answer to inverse oflaplace 1/(s-1(s^2)) is \H(t)-e^t