What is inverse oflaplace 1/(0.012s^2+1.55s+50) ?

The answer to inverse oflaplace 1/(0.012s^2+1.55s+50) is 40.00000…e^{-64.58333…t}sinh(2.08333…t)

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inverse oflaplace 1/(0.012s^2+1.55s+50)

inverse laplace

\frac{1}{0.012 s ^{2} + 1.55 s + 50}

40.00000 \dots e^{- 64.58333 \dots t} sinh (2.08333 \dots t)

Solution steps

L^{- 1} {\frac{1}{0.012 s ^{2} + 1.55 s + 50}}

Expand

\frac{1}{0.012 s ^{2} + 1.55 s + 50} : 83.33333 \dots \cdot \frac{1}{( s + 64.58333 \dots ) ^{2} - 4.34027 \dots}

= L^{- 1} {83.33333 \dots \cdot \frac{1}{( s + 64.58333 \dots ) ^{2} - 4.34027 \dots}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= 83.33333 \dots L^{- 1} {\frac{1}{( s + 64.58333 \dots ) ^{2} - 4.34027 \dots}}

L^{- 1} {\frac{1}{( s + 64.58333 \dots ) ^{2} - 4.34027 \dots}} : e^{- 64.58333 \dots t} \frac{1}{2.08333 \dots} sinh (2.08333 \dots t)

= 83.33333 \dots e^{- 64.58333 \dots t} \frac{1}{2.08333 \dots} sinh (2.08333 \dots t)

Simplify

83.33333 \dots e^{- 64.58333 \dots t} \frac{1}{2.08333 \dots} sinh (2.08333 \dots t) : 40.00000 \dots e^{- 64.58333 \dots t} sinh (2.08333 \dots t)

= 40.00000 \dots e^{- 64.58333 \dots t} sinh (2.08333 \dots t)

s \to 1 lim (\frac{( 2 s + 3 )}{s})

\frac{d y}{d x} = 2 y + 3

\int_{1}^{8} (x^{2} - 4 x + 6) d x

f (x) = ln (x - x^{2})

\int 2 x^{2} - 3 x d x

What is inverse oflaplace 1/(0.012s^2+1.55s+50) ?
The answer to inverse oflaplace 1/(0.012s^2+1.55s+50) is 40.00000…e^{-64.58333…t}sinh(2.08333…t)