What is inverse oflaplace (3s+1)/(s(s+1)(-s+5)) ?

The answer to inverse oflaplace (3s+1)/(s(s+1)(-s+5)) is 1/5 \H(t)+1/3 e^{-t}-8/15 e^{5t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (3s+1)/(s(s+1)(-s+5))

inverse laplace

\frac{3 s + 1}{s ( s + 1 ) ( - s + 5 )}

\frac{1}{5} H (t) + \frac{1}{3} e^{- t} - \frac{8}{15} e^{5 t}

Solution steps

L^{- 1} {\frac{3 s + 1}{s ( s + 1 ) ( - s + 5 )}}

Take the partial fraction of

\frac{3 s + 1}{s ( s + 1 ) ( - s + 5 )} : \frac{1}{5 s} + \frac{1}{3 ( s + 1 )} - \frac{8}{15 ( s - 5 )}

= L^{- 1} {\frac{1}{5 s} + \frac{1}{3 ( s + 1 )} - \frac{8}{15 ( s - 5 )}}

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}{5 s}} + L^{- 1} {\frac{1}{3 ( s + 1 )}} - L^{- 1} {\frac{8}{15 ( s - 5 )}}

L^{- 1} {\frac{1}{5 s}} : \frac{1}{5} H (t)

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

L^{- 1} {\frac{8}{15 ( s - 5 )}} : \frac{8}{15} e^{5 t}

= \frac{1}{5} H (t) + \frac{1}{3} e^{- t} - \frac{8}{15} e^{5 t}

\int_{0}^{1} \frac{1}{x - 3} d x

y^{^{''}} - 81 y = 0, y (2) = 2, y (- 1) = - 1

\frac{d}{d x} (x + e)

\frac{1}{sin ( x ) cos ( x + b + c )}

\int_{0}^{1} arcsin (x) - sin (x) d x

What is inverse oflaplace (3s+1)/(s(s+1)(-s+5)) ?
The answer to inverse oflaplace (3s+1)/(s(s+1)(-s+5)) is 1/5 \H(t)+1/3 e^{-t}-8/15 e^{5t}