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

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

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inverse oflaplace (s+1)/((s-1)*(s-2))

inverse laplace

\frac{s + 1}{( s - 1 ) \cdot ( s - 2 )}

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

Solution steps

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

Take the partial fraction of

\frac{s + 1}{( s - 1 ) ( 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}

tan (θ) sec (θ)

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

f (x) = \frac{x}{x - 1}, at x = 2

\frac{d}{d x} ((x + 1)^{5})

\frac{d}{d x} (\frac{2}{5} x^{10})

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