What is inverse oflaplace (s^3+5s^2+9s+7)/(s^2+3s+2) ?

The answer to inverse oflaplace (s^3+5s^2+9s+7)/(s^2+3s+2) is \delta'(t)+2\delta(t)+2e^{-t}-e^{-2t}

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inverse oflaplace (s^3+5s^2+9s+7)/(s^2+3s+2)

inverse laplace

\frac{s ^{3} + 5 s ^{2} + 9 s + 7}{s ^{2} + 3 s + 2}

δ' (t) + 2 δ (t) + 2 e^{- t} - e^{- 2 t}

Solution steps

L^{- 1} {\frac{s ^{3} + 5 s ^{2} + 9 s + 7}{s ^{2} + 3 s + 2}}

Take the partial fraction of

\frac{s ^{3} + 5 s ^{2} + 9 s + 7}{s ^{2} + 3 s + 2} : s + 2 + \frac{2}{s + 1} - \frac{1}{s + 2}

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

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {2} : 2 δ (t)

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

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

= δ^{^{'}} (t) + 2 δ (t) + 2 e^{- t} - e^{- 2 t}

\frac{\partial}{\partial x} (a x^{2})

x \to - 3 lim (\frac{x ^{2} - 4}{3 - x})

\frac{x - 4}{x + 2}

\int \frac{2 x + 1}{( x - 1 ) ( x + 4 ) ^{2}} d x

\int \frac{1}{x - 1 - x} d x

What is inverse oflaplace (s^3+5s^2+9s+7)/(s^2+3s+2) ?
The answer to inverse oflaplace (s^3+5s^2+9s+7)/(s^2+3s+2) is \delta'(t)+2\delta(t)+2e^{-t}-e^{-2t}