What is inverse oflaplace 1/((s-1)^2-4) ?

The answer to inverse oflaplace 1/((s-1)^2-4) is -1/4 e^{-t}+1/4 e^{3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 1/((s-1)^2-4)

inverse laplace

\frac{1}{( s - 1 ) ^{2} - 4}

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

Solution steps

L^{- 1} {\frac{1}{( s - 1 ) ^{2} - 4}}

Take the partial fraction of

\frac{1}{( s - 1 ) ^{2} - 4} : - \frac{1}{4 ( s + 1 )} + \frac{1}{4 ( s - 3 )}

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

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

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

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

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

(x^{2} + 2) y^{^{'}} + 2 x y = x^{3}

\int_{0}^{5} 4 x^{2} d x

\int 2 z^{2} d z

\int \frac{1}{3 - 6 x - x ^{2}} d x

x \to 0 lim (5 x + 4)

What is inverse oflaplace 1/((s-1)^2-4) ?
The answer to inverse oflaplace 1/((s-1)^2-4) is -1/4 e^{-t}+1/4 e^{3t}