What is inverse oflaplace ((s+1)^2)/((s+2)^4) ?

The answer to inverse oflaplace ((s+1)^2)/((s+2)^4) is e^{-2t}t-e^{-2t}t^2+(e^{-2t}t^3)/6

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inverse oflaplace ((s+1)^2)/((s+2)^4)

inverse laplace

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

e^{- 2 t} t - e^{- 2 t} t^{2} + \frac{e ^{- 2 t} t ^{3}}{6}

Solution steps

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

Take the partial fraction of

\frac{( s + 1 ) ^{2}}{( s + 2 ) ^{4}} : \frac{1}{( s + 2 ) ^{2}} - \frac{2}{( s + 2 ) ^{3}} + \frac{1}{( s + 2 ) ^{4}}

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

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

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

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

L^{- 1} {\frac{1}{( s + 2 ) ^{4}}} : \frac{e ^{- 2 t} t ^{3}}{6}

= e^{- 2 t} t - e^{- 2 t} t^{2} + \frac{e ^{- 2 t} t ^{3}}{6}

y = \frac{3}{x}, (4, \frac{3}{4})

\int \frac{( 2 x + 1 )}{x - 1} d x

y^{^{''}} + 3 y^{^{'}} - 18 y = 0

\frac{d}{d x} (\frac{2 sin ( x )}{3 + cos ( x )})

y = 96 x^{2} + 72 x + 12

What is inverse oflaplace ((s+1)^2)/((s+2)^4) ?
The answer to inverse oflaplace ((s+1)^2)/((s+2)^4) is e^{-2t}t-e^{-2t}t^2+(e^{-2t}t^3)/6