What is inverse oflaplace (s-3)/((s-4)(s+6)) ?

The answer to inverse oflaplace (s-3)/((s-4)(s+6)) is 1/10 e^{4t}+9/10 e^{-6t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (s-3)/((s-4)(s+6))

inverse laplace

\frac{s - 3}{( s - 4 ) ( s + 6 )}

\frac{1}{10} e^{4 t} + \frac{9}{10} e^{- 6 t}

Solution steps

L^{- 1} {\frac{s - 3}{( s - 4 ) ( s + 6 )}}

Take the partial fraction of

\frac{s - 3}{( s - 4 ) ( s + 6 )} : \frac{1}{10 ( s - 4 )} + \frac{9}{10 ( s + 6 )}

= L^{- 1} {\frac{1}{10 ( s - 4 )} + \frac{9}{10 ( s + 6 )}}

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{1}{10 ( s - 4 )}} + L^{- 1} {\frac{9}{10 ( s + 6 )}}

L^{- 1} {\frac{1}{10 ( s - 4 )}} : \frac{1}{10} e^{4 t}

L^{- 1} {\frac{9}{10 ( s + 6 )}} : \frac{9}{10} e^{- 6 t}

= \frac{1}{10} e^{4 t} + \frac{9}{10} e^{- 6 t}

\int \frac{x + 1}{x ^{3} + 7 x ^{2} + 6 x} d x

\frac{d}{d x} (\frac{4}{5 x ^{6}})

f (x) = 2 - 6 x^{2}, (- 5, - 148)

\frac{d}{d x} ((3 x^{4} - 6 x)^{3})

y = 4 e^{x} cos (x)

What is inverse oflaplace (s-3)/((s-4)(s+6)) ?
The answer to inverse oflaplace (s-3)/((s-4)(s+6)) is 1/10 e^{4t}+9/10 e^{-6t}