What is inverse oflaplace s/((s+4)^2) ?

The answer to inverse oflaplace s/((s+4)^2) is e^{-4t}-e^{-4t}*4t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace s/((s+4)^2)

inverse laplace

\frac{s}{( s + 4 ) ^{2}}

e^{- 4 t} - e^{- 4 t} \cdot 4 t

Solution steps

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

Take the partial fraction of

\frac{s}{( s + 4 ) ^{2}} : \frac{1}{s + 4} - \frac{4}{( s + 4 ) ^{2}}

= L^{- 1} {\frac{1}{s + 4} - \frac{4}{( s + 4 ) ^{2}}}

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

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

L^{- 1} {\frac{4}{( s + 4 ) ^{2}}} : e^{- 4 t} \cdot 4 t

= e^{- 4 t} - e^{- 4 t} \cdot 4 t

n = 1 \sum \infty \frac{9 n}{7 n + 1}

y = x, y = 0, y = x - 2

x \to 3 + lim (\frac{2}{x - 3})

y^{^{'}} = y - 4 y^{2}

y^{^{'}} = \frac{x ^{2}}{y ( 1 + x ^{3} )}

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