What is inverse oflaplace 4s-1 ?

The answer to inverse oflaplace 4s-1 is 4\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 4s-1

inverse laplace

4 s - 1

4 δ' (t) - δ (t)

Solution steps

L^{- 1} {4 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)}

= 4 L^{- 1} {s} - L^{- 1} {1}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {1} : δ (t)

= 4 δ^{^{'}} (t) - δ (t)

f (x) = 2 x^{2} + 4 x, (- 3, 6)

x \to 1 lim (ln (x^{5} - 1))

\int_{0}^{1} π (x - x^{12}) d x

\int \frac{3 x ^{2} - x + 2 e}{x ^{2}} d x

\int_{2}^{3} π (x + x^{2})^{2} d x

What is inverse oflaplace 4s-1 ?
The answer to inverse oflaplace 4s-1 is 4\delta'(t)-\delta(t)