What is inverse oflaplace 3/(s^2-4) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 3/(s^2-4)

inverse laplace

\frac{3}{s ^{2} - 4}

- \frac{3}{4} e^{- 2 t} + \frac{3}{4} e^{2 t}

Solution steps

L^{- 1} {\frac{3}{s ^{2} - 4}}

Take the partial fraction of

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

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

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

L^{- 1} {\frac{3}{4 ( s - 2 )}} : \frac{3}{4} e^{2 t}

= - \frac{3}{4} e^{- 2 t} + \frac{3}{4} e^{2 t}

\frac{d}{d x} (x^{3 l n (x)})

\int - \frac{2}{3} y^{- \frac{4}{3}} d y

y = \frac{3 cos ( x )}{1 + sin ( x )}

\frac{d}{d x} ((tan (x))^{\frac{1}{2}})

\int_{0}^{l n (3)} e^{x} d x

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