What is inverse oflaplace-3/(s+3)+4/(s-4) ?

The answer to inverse oflaplace-3/(s+3)+4/(s-4) is -3e^{-3t}+4e^{4t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace-3/(s+3)+4/(s-4)

inverse laplace

- \frac{3}{s + 3} + \frac{4}{s - 4}

- 3 e^{- 3 t} + 4 e^{4 t}

Solution steps

L^{- 1} {- \frac{3}{s + 3} + \frac{4}{s - 4}}

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} {\frac{3}{s + 3}} + L^{- 1} {\frac{4}{s - 4}}

L^{- 1} {\frac{3}{s + 3}} : 3 e^{- 3 t}

L^{- 1} {\frac{4}{s - 4}} : 4 e^{4 t}

= - 3 e^{- 3 t} + 4 e^{4 t}

\int \frac{8}{x ^{- 1} + 2} d x

y = 4 x^{2} - x

\frac{e ^{x} + 1}{e ^{2 x}}

\frac{( e ^{2 x} - 1 )}{x}, - 0.5

\frac{d y}{d x} = (\frac{sec ( x )}{y})^{2}

What is inverse oflaplace-3/(s+3)+4/(s-4) ?
The answer to inverse oflaplace-3/(s+3)+4/(s-4) is -3e^{-3t}+4e^{4t}