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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

\frac{s + 1}{s + 2}

δ (t) - e^{- 2 t}

Solution steps

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

Take the partial fraction of

\frac{s + 1}{s + 2} : 1 - \frac{1}{s + 2}

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

L^{- 1} {1} : δ (t)

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

= δ (t) - e^{- 2 t}

\frac{d}{d x} (x^{2} + 4 x^{3})

\frac{d}{d x} (- 25 sin (5 x))

f (x) = c x^{- 2}

\frac{\partial}{\partial x} (e^{x} cos (x^{2} + y^{2}))

y = x^{4} + 6 e^{x}, (0, 6)

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