What is inverse oflaplace s+1 ?

The answer to inverse oflaplace s+1 is \delta'(t)+\delta(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace s+1

inverse laplace

s + 1

δ' (t) + δ (t)

Solution steps

L^{- 1} {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} {s} + L^{- 1} {1}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {1} : δ (t)

= δ^{^{'}} (t) + δ (t)

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

\frac{d}{d x} (sin (e^{x^{5} s i n (x)}))

y^{^{'}} = 4.9 - 22.22 y

(x - 5) (x + 4) (x + 5)

\int tan^{3} (x) sec^{7} (x) d x

What is inverse oflaplace s+1 ?
The answer to inverse oflaplace s+1 is \delta'(t)+\delta(t)