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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 4/(s^2(s+2))

inverse laplace

\frac{4}{s ^{2} ( s + 2 )}

- H (t) + 2 t + e^{- 2 t}

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {- \frac{1}{s} + \frac{2}{s ^{2}} + \frac{1}{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}} + L^{- 1} {\frac{2}{s ^{2}}} + L^{- 1} {\frac{1}{s + 2}}

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

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

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

= - H (t) + 2 t + e^{- 2 t}

y^{^{''}} - 4 y = 16 x e^{2 x}

\frac{d y}{d t} + tan (t) y = 3 sin (t), y (0) = 1

\frac{d y}{d x} = \frac{3 x ^{2}}{y ( 1 - x ^{3} )}

\frac{d}{d x} (- 32 cos (8 x))

\frac{d}{d x} (\frac{2 x + 9}{2 x})

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