What is inverse oflaplace s/((s-3)(s-6)(s-2)) ?

The answer to inverse oflaplace s/((s-3)(s-6)(s-2)) is -e^{3t}+1/2 e^{6t}+1/2 e^{2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace s/((s-3)(s-6)(s-2))

inverse laplace

\frac{s}{( s - 3 ) ( s - 6 ) ( s - 2 )}

- e^{3 t} + \frac{1}{2} e^{6 t} + \frac{1}{2} e^{2 t}

Solution steps

L^{- 1} {\frac{s}{( s - 3 ) ( s - 6 ) ( s - 2 )}}

Take the partial fraction of

\frac{s}{( s - 3 ) ( s - 6 ) ( s - 2 )} : - \frac{1}{s - 3} + \frac{1}{2 ( s - 6 )} + \frac{1}{2 ( s - 2 )}

= L^{- 1} {- \frac{1}{s - 3} + \frac{1}{2 ( s - 6 )} + \frac{1}{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{1}{s - 3}} + L^{- 1} {\frac{1}{2 ( s - 6 )}} + L^{- 1} {\frac{1}{2 ( s - 2 )}}

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

L^{- 1} {\frac{1}{2 ( s - 6 )}} : \frac{1}{2} e^{6 t}

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

= - e^{3 t} + \frac{1}{2} e^{6 t} + \frac{1}{2} e^{2 t}

\int_{0}^{0.4} \frac{( x ^{2} )}{4 - 25 x ^{2}} d x

\int (sin (x)) (cos (x)) d x

\frac{d}{d x} ((x^{5} - 9) x)

x = - 3, x = 1, y = x^{3} + 1, y = 0

\int_{2}^{\infty} x e^{- 2 x} d x

What is inverse oflaplace s/((s-3)(s-6)(s-2)) ?
The answer to inverse oflaplace s/((s-3)(s-6)(s-2)) is -e^{3t}+1/2 e^{6t}+1/2 e^{2t}