What is inverse oflaplace e^{-2s}(1/(s+1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace e^{-2s}(1/(s+1))

inverse laplace

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

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

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

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

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

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

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

\int t e^{- 2 t} d t

\frac{\partial}{\partial x} (50 x - 0.5 x^{2} + 75 y - y^{2})

\frac{\partial}{\partial x} (- ln (x^{2} + 1))

n = 1 \sum \infty (x + 6)^{n}

\int e^{- x^{3}} d x

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