What is inverse oflaplace (s^2)/(s^2+20s+100) ?

The answer to inverse oflaplace (s^2)/(s^2+20s+100) is \delta(t)-20e^{-10t}+e^{-10t}*100t

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inverse oflaplace (s^2)/(s^2+20s+100)

inverse laplace

\frac{s ^{2}}{s ^{2} + 20 s + 100}

δ (t) - 20 e^{- 10 t} + e^{- 10 t} \cdot 100 t

Solution steps

L^{- 1} {\frac{s ^{2}}{s ^{2} + 20 s + 100}}

Take the partial fraction of

\frac{s ^{2}}{s ^{2} + 20 s + 100} : 1 - \frac{20}{s + 10} + \frac{100}{( s + 10 ) ^{2}}

= L^{- 1} {1 - \frac{20}{s + 10} + \frac{100}{( s + 10 ) ^{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} {1} - L^{- 1} {\frac{20}{s + 10}} + L^{- 1} {\frac{100}{( s + 10 ) ^{2}}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{20}{s + 10}} : 20 e^{- 10 t}

L^{- 1} {\frac{100}{( s + 10 ) ^{2}}} : e^{- 10 t} \cdot 100 t

= δ (t) - 20 e^{- 10 t} + e^{- 10 t} \cdot 100 t

x \to - 1 lim (x^{2} - 2)

\int - 4 sin (2 x) d x

\frac{d}{d x} (\frac{3 + 8 x}{5 - 4 x})

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

\frac{\partial}{\partial y} (x cos (y) sin (z))

What is inverse oflaplace (s^2)/(s^2+20s+100) ?
The answer to inverse oflaplace (s^2)/(s^2+20s+100) is \delta(t)-20e^{-10t}+e^{-10t}*100t