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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

\frac{4}{3} e^{2 t} - \frac{1}{3} e^{- t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= \frac{4}{3} e^{2 t} - \frac{1}{3} e^{- t}

\int 2 x y - sin (x) d x

f (x) = (x + 1) (3 x^{2} - 5 x + 2)

y^{^{''}} - y^{^{'}} - y = 0

x \to 4 lim (2 x - 1)

\frac{d}{d x} (2 e^{x} ln (x))

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