What is inverse oflaplace (5(s+2)^2)/(s(s+1)^3) ?

The answer to inverse oflaplace (5(s+2)^2)/(s(s+1)^3) is 20\H(t)-20e^{-t}-e^{-t}*15t-(5e^{-t}t^2)/2

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inverse oflaplace (5(s+2)^2)/(s(s+1)^3)

inverse laplace

\frac{5 ( s + 2 ) ^{2}}{s ( s + 1 ) ^{3}}

20 H (t) - 20 e^{- t} - e^{- t} \cdot 15 t - \frac{5 e ^{- t} t ^{2}}{2}

Solution steps

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

Take the partial fraction of

\frac{5 ( s + 2 ) ^{2}}{s ( s + 1 ) ^{3}} : \frac{20}{s} - \frac{20}{s + 1} - \frac{15}{( s + 1 ) ^{2}} - \frac{5}{( s + 1 ) ^{3}}

= L^{- 1} {\frac{20}{s} - \frac{20}{s + 1} - \frac{15}{( s + 1 ) ^{2}} - \frac{5}{( s + 1 ) ^{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{20}{s}} - L^{- 1} {\frac{20}{s + 1}} - L^{- 1} {\frac{15}{( s + 1 ) ^{2}}} - L^{- 1} {\frac{5}{( s + 1 ) ^{3}}}

L^{- 1} {\frac{20}{s}} : 20 H (t)

L^{- 1} {\frac{20}{s + 1}} : 20 e^{- t}

L^{- 1} {\frac{15}{( s + 1 ) ^{2}}} : e^{- t} \cdot 15 t

L^{- 1} {\frac{5}{( s + 1 ) ^{3}}} : \frac{5 e ^{- t} t ^{2}}{2}

= 20 H (t) - 20 e^{- t} - e^{- t} \cdot 15 t - \frac{5 e ^{- t} t ^{2}}{2}

\int (- 25 sin (5 x - \frac{π}{2})) d x

\int \frac{4}{( 1 + x ^{2} )} d x

f (x) = (x - 3)^{2} - 2, at x = 0

\frac{1}{x} y^{^{'}} = \frac{y}{x ^{2}} - \frac{1}{y}

\frac{\partial}{\partial y} (y^{3} + x^{2} y - 3 y)

What is inverse oflaplace (5(s+2)^2)/(s(s+1)^3) ?
The answer to inverse oflaplace (5(s+2)^2)/(s(s+1)^3) is 20\H(t)-20e^{-t}-e^{-t}*15t-(5e^{-t}t^2)/2