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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

\frac{5}{3} H (t) - \frac{5}{3} e^{- 3 t}

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{5}{3 s}} : \frac{5}{3} H (t)

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

= \frac{5}{3} H (t) - \frac{5}{3} e^{- 3 t}

\int (cos (x)) (sin (x)) d x

\frac{d}{d x} (\frac{4 x ^{2} + 8 x + 2}{x})

x^{2} \frac{d y}{d x} + sin (x) - y = 0

x \to + 0 lim (3 x^{2})

\frac{d y}{d x} = y sin (2 x)

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