What is inverse oflaplace 1/(s^2+10.1s+1) ?

The answer to inverse oflaplace 1/(s^2+10.1s+1) is e^{-5.05t} 1/(4.95)sinh(4.95t)

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inverse oflaplace 1/(s^2+10.1s+1)

inverse laplace

\frac{1}{s ^{2} + 10.1 s + 1}

e^{- 5.05 t} \frac{1}{4.95} sinh (4.95 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 10.1 s + 1}}

\frac{1}{s ^{2} + 10.1 s + 1} = \frac{1}{( s + 5.05 ) ^{2} - 24.5025}

= L^{- 1} {\frac{1}{( s + 5.05 ) ^{2} - 24.5025}}

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 + 5.05 ) ^{2} - 24.5025} : a = - 5.05, F (s) = \frac{1}{s ^{2} - 24.5025}

= e^{- 5.05 t} L^{- 1} {\frac{1}{s ^{2} - 24.5025}}

L^{- 1} {\frac{1}{s ^{2} - 24.5025}} : \frac{1}{4.95} sinh (4.95 t)

= e^{- 5.05 t} \frac{1}{4.95} sinh (4.95 t)

\int x - 2 d x

f (x) = 3 x + x^{\frac{4}{3}}

x \to 0 + lim (\frac{1}{x} \cdot e^{\frac{1}{x}})

f (x) = 3 cos (2 x)

y = \frac{8}{1 + x ^{4}}, y = 4 x^{2}

What is inverse oflaplace 1/(s^2+10.1s+1) ?
The answer to inverse oflaplace 1/(s^2+10.1s+1) is e^{-5.05t} 1/(4.95)sinh(4.95t)