What is inverse oflaplace 89/2 s ?

The answer to inverse oflaplace 89/2 s is 89/2 \delta'(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 89/2 s

inverse laplace

\frac{89}{2} s

\frac{89}{2} δ' (t)

Solution steps

L^{- 1} {\frac{89}{2} s}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{89}{2} L^{- 1} {s}

L^{- 1} {s} : δ^{'} (t)

= \frac{89}{2} δ^{^{'}} (t)

(e^{- \frac{1}{x}})^{^{'}}

f (x) = (x^{2} + 2 x)^{3} (7 x + 1)^{4}

\frac{d y}{d x} = - \frac{x}{y}

\frac{d}{d x} (2 sin (x) + 4 cos (x) sin (x))

\int \frac{1}{( x ^{2} ) \cdot 25 x ^{2} + 16} d x

What is inverse oflaplace 89/2 s ?
The answer to inverse oflaplace 89/2 s is 89/2 \delta'(t)