What is inverse oflaplace-4e^{(-2s)}*1/(s-3) ?

The answer to inverse oflaplace-4e^{(-2s)}*1/(s-3) is -4\H(t-2)e^{3(t-2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace-4e^{(-2s)}*1/(s-3)

inverse laplace

- 4 e^{(- 2 s)} \cdot \frac{1}{s - 3}

- 4 H (t - 2) e^{3 (t - 2)}

Solution steps

L^{- 1} {- 4 e^{(- 2 s)} \frac{1}{s - 3}}

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)}

= - 4 L^{- 1} {e^{- 2 s} \frac{1}{s - 3}}

L^{- 1} {e^{- 2 s} \frac{1}{s - 3}} : H (t - 2) e^{3 (t - 2)}

= - 4 H (t - 2) e^{3 (t - 2)}

x^{(3)} + a \cdot x^{(2)} + b \cdot x + c

\frac{d}{d x} (2 x - 9)

\int (x - 1)^{\frac{1}{2}} d x

\frac{x}{5} + \frac{x ^{2} - 100 x + 6100}{1.1}

\int sin (y) cosh (x) d y

What is inverse oflaplace-4e^{(-2s)}*1/(s-3) ?
The answer to inverse oflaplace-4e^{(-2s)}*1/(s-3) is -4\H(t-2)e^{3(t-2)}