What is inverse oflaplace (20)/(s(s^2+4s+4)) ?

The answer to inverse oflaplace (20)/(s(s^2+4s+4)) is 5\H(t)-5e^{-2t}-e^{-2t}*10t

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inverse oflaplace (20)/(s(s^2+4s+4))

inverse laplace

\frac{20}{s ( s ^{2} + 4 s + 4 )}

5 H (t) - 5 e^{- 2 t} - e^{- 2 t} \cdot 10 t

Solution steps

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

Take the partial fraction of

\frac{20}{s ( s ^{2} + 4 s + 4 )} : \frac{5}{s} - \frac{5}{s + 2} - \frac{10}{( s + 2 ) ^{2}}

= L^{- 1} {\frac{5}{s} - \frac{5}{s + 2} - \frac{10}{( s + 2 ) ^{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{5}{s}} - L^{- 1} {\frac{5}{s + 2}} - L^{- 1} {\frac{10}{( s + 2 ) ^{2}}}

L^{- 1} {\frac{5}{s}} : 5 H (t)

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

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

= 5 H (t) - 5 e^{- 2 t} - e^{- 2 t} \cdot 10 t

x \to \infty lim (\frac{7 x ^{3} + 8 x}{4 x - 3})

\int (\frac{W}{U}) d W

\frac{d}{d x} (2 x - x^{3})

\frac{d}{d x} (x^{2} + y^{2} + z^{2} + x y)

x y^{2} + 3 y^{2} - x^{2} y^{^{'}} = 0

What is inverse oflaplace (20)/(s(s^2+4s+4)) ?
The answer to inverse oflaplace (20)/(s(s^2+4s+4)) is 5\H(t)-5e^{-2t}-e^{-2t}*10t