What is inverse oflaplace 1/(s^2+0.1s+1) ?

The answer to inverse oflaplace 1/(s^2+0.1s+1) is 1.00125…e^{-0.05t}sin(0.99874…t)

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inverse oflaplace 1/(s^2+0.1s+1)

inverse laplace

\frac{1}{s ^{2} + 0.1 s + 1}

1.00125 \dots e^{- 0.05 t} sin (0.99874 \dots t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 0.1 s + 1}}

\frac{1}{s ^{2} + 0.1 s + 1} = \frac{1}{( s + 0.05 ) ^{2} + 0.9975}

= L^{- 1} {\frac{1}{( s + 0.05 ) ^{2} + 0.9975}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s + 0.05 ) ^{2} + 0.9975} : a = - 0.05, F (s) = \frac{1}{s ^{2} + 0.9975}

= e^{- 0.05 t} L^{- 1} {\frac{1}{s ^{2} + 0.9975}}

L^{- 1} {\frac{1}{s ^{2} + 0.9975}} : \frac{1}{0.99874 \dots} sin (0.99874 \dots t)

Simplify

e^{- 0.05 t} \frac{1}{0.99874 \dots} sin (0.99874 \dots t) : 1.00125 \dots e^{- 0.05 t} sin (0.99874 \dots t)

= 1.00125 \dots e^{- 0.05 t} sin (0.99874 \dots t)

\int (n x)^{\frac{( n - 1 )}{n}} d x

x \to 0 lim (x (x - [x]))

\int x cos (\frac{nπ x}{5}) d x

(1 + x)^{\frac{1}{6}}

\int_{0}^{2 π} ∣ sin (x) ∣ d x

What is inverse oflaplace 1/(s^2+0.1s+1) ?
The answer to inverse oflaplace 1/(s^2+0.1s+1) is 1.00125…e^{-0.05t}sin(0.99874…t)