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

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

Popular >

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

\frac{d}{d x} (6 arcsin (x^{3}))

\frac{d}{d x} ((1 + cos (2 x))^{2})

\frac{d}{d x} (\frac{x ^{2} - 2 x}{x})

f (x) = x^{2} + 3 x, (1, 2)

\int \frac{9 - 9 x}{1 - x} d x

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