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

The answer to inverse oflaplace 1/((s-3)(s+1)) is 1/4 e^{3t}-1/4 e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

\frac{1}{( s - 3 ) ( s + 1 )}

\frac{1}{4} e^{3 t} - \frac{1}{4} e^{- t}

Solution steps

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

Take the partial fraction of

\frac{1}{( s - 3 ) ( s + 1 )} : \frac{1}{4 ( s - 3 )} - \frac{1}{4 ( s + 1 )}

= L^{- 1} {\frac{1}{4 ( s - 3 )} - \frac{1}{4 ( s + 1 )}}

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}{4 ( s - 3 )}} - L^{- 1} {\frac{1}{4 ( s + 1 )}}

L^{- 1} {\frac{1}{4 ( s - 3 )}} : \frac{1}{4} e^{3 t}

L^{- 1} {\frac{1}{4 ( s + 1 )}} : \frac{1}{4} e^{- t}

= \frac{1}{4} e^{3 t} - \frac{1}{4} e^{- t}

\int \frac{3 x - y}{9} d y

y^{^{''}} - 14 y^{^{'}} + 49 y = 3 t^{- 1} e^{7 t}

\frac{d}{d x} (2 csc (x) + 4 cot (x))

(- 3, 4), (2, - 9)

\int tan^{- \frac{3}{2}} (x) sec^{4} (x) d x

What is inverse oflaplace 1/((s-3)(s+1)) ?
The answer to inverse oflaplace 1/((s-3)(s+1)) is 1/4 e^{3t}-1/4 e^{-t}