What is inverse oflaplace (5s)/((s+4)^3) ?

The answer to inverse oflaplace (5s)/((s+4)^3) is e^{-4t}*5t-e^{-4t}*10t^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (5s)/((s+4)^3)

inverse laplace

\frac{5 s}{( s + 4 ) ^{3}}

e^{- 4 t} \cdot 5 t - e^{- 4 t} \cdot 10 t^{2}

Solution steps

L^{- 1} {\frac{5 s}{( s + 4 ) ^{3}}}

Take the partial fraction of

\frac{5 s}{( s + 4 ) ^{3}} : \frac{5}{( s + 4 ) ^{2}} - \frac{20}{( s + 4 ) ^{3}}

= L^{- 1} {\frac{5}{( s + 4 ) ^{2}} - \frac{20}{( s + 4 ) ^{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}{( s + 4 ) ^{2}}} - L^{- 1} {\frac{20}{( s + 4 ) ^{3}}}

L^{- 1} {\frac{5}{( s + 4 ) ^{2}}} : e^{- 4 t} \cdot 5 t

L^{- 1} {\frac{20}{( s + 4 ) ^{3}}} : e^{- 4 t} \cdot 10 t^{2}

= e^{- 4 t} \cdot 5 t - e^{- 4 t} \cdot 10 t^{2}

\frac{\partial}{\partial x} (\frac{2 x}{( x ^{2} + y ^{2} )})

(2 x + 5) (5 x^{2} + 1)

\int_{0}^{\frac{3 π}{2}} 7 ∣ sin (x) ∣ d x

f (x) = e^{x}, g (x) = cos (x), x = 0, x = \frac{π}{2}

y = x^{2} - 3 x, y = x

What is inverse oflaplace (5s)/((s+4)^3) ?
The answer to inverse oflaplace (5s)/((s+4)^3) is e^{-4t}*5t-e^{-4t}*10t^2