What is inverse oflaplace e^{-2s} 1/(s-9) ?

The answer to inverse oflaplace e^{-2s} 1/(s-9) is \H(t-2)e^{9(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-9)

inverse laplace

e^{- 2 s} \frac{1}{s - 9}

H (t - 2) e^{9 (t - 2)}

Solution steps

L^{- 1} {e^{- 2 s} \frac{1}{s - 9}}

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 - 9}} (t - 2)

L^{- 1} {\frac{1}{s - 9}} : e^{9 t}

= H (t - 2) e^{9 (t - 2)}

\int \frac{1}{1 + x} d x

(0, - 6), (3, - 5)

t \to 0 lim (\frac{sin ( 3 t ) + 4 t}{t sec ( t )})

\frac{\partial}{\partial y} (5 x ln (x y))

What is inverse oflaplace e^{-2s} 1/(s-9) ?
The answer to inverse oflaplace e^{-2s} 1/(s-9) is \H(t-2)e^{9(t-2)}