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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\frac{\partial}{\partial x} (9 x + y + 7 x y)

f^{^{'}} (x) = 1 - 2 x

x \to \infty lim (x^{2} + 5 x)

\frac{d}{d x} (3 x + 6)

\int (x^{8} - 16)^{9} x^{7} d x

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