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

The answer to inverse oflaplace-(2s+1)/(s^2+3s+2) is e^{-t}-3e^{-2t}

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

inverse laplace

- \frac{2 s + 1}{s ^{2} + 3 s + 2}

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

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{3}{s + 2}} : 3 e^{- 2 t}

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

\frac{\partial}{\partial y} (x^{3} y^{6})

\int (3 x^{- 2} - 4 x^{2} + 1) d x

\frac{\partial}{\partial x} (e^{- 9 x} cos (θ))

\frac{s}{s ^{2} + 4 ^{2}}

\int \frac{e ^{x} + 9}{e ^{2 x} - 6 e ^{x} + 9} d x

What is inverse oflaplace-(2s+1)/(s^2+3s+2) ?
The answer to inverse oflaplace-(2s+1)/(s^2+3s+2) is e^{-t}-3e^{-2t}