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

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

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

inverse laplace

\frac{( 2 s - 3 )}{( s ^{2} - 3 s + 2 )}

e^{t} + e^{2 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= e^{t} + e^{2 t}

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