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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{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}

arctan (5 x^{2} - 1)

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} = 1 + 9 x^{2}

\int 3 sec^{4} (x) d x

x \to \frac{3 π}{4} lim (- 2 sec (x))

\frac{d x}{d t} = 6 \cdot 0.3 - 5 \cdot \frac{x}{100 + t}, x (0) = 0

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