What is inverse oflaplace (s^2+6s+2)/(s(s+1)^2) ?

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

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inverse oflaplace (s^2+6s+2)/(s(s+1)^2)

inverse laplace

\frac{s ^{2} + 6 s + 2}{s ( s + 1 ) ^{2}}

2 H (t) - e^{- t} + e^{- t} \cdot 3 t

Solution steps

L^{- 1} {\frac{s ^{2} + 6 s + 2}{s ( s + 1 ) ^{2}}}

Take the partial fraction of

\frac{s ^{2} + 6 s + 2}{s ( s + 1 ) ^{2}} : \frac{2}{s} - \frac{1}{s + 1} + \frac{3}{( s + 1 ) ^{2}}

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

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

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

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

= 2 H (t) - e^{- t} + e^{- t} \cdot 3 t

y^{^{'}} = t y - 2 t + y - 2

3 t^{3} - 2 t^{2} - 7 t + 2

\frac{\partial}{\partial x} (1^{x})

\frac{( x ^{2} + 11 )}{x ^{3}}

(x^{''} + 4 x)^{^{'}}

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