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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

x \to 0 lim (\frac{cos ( x ^{2} )}{x})

y^{^{''}} - 4 y^{^{'}} - 5 y = 50 t^{3}

\int \frac{e ^{- x} - e ^{2 x}}{2} d x

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

\frac{d}{d x} (\frac{m}{x})

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