What is inverse oflaplace 1/(7s^2+37.4s+19990) ?

The answer to inverse oflaplace 1/(7s^2+37.4s+19990) is 0.00267…e^{-2.67142…t}sin(53.37206…t)

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inverse oflaplace 1/(7s^2+37.4s+19990)

inverse laplace

\frac{1}{7 s ^{2} + 37.4 s + 19990}

0.00267 \dots e^{- 2.67142 \dots t} sin (53.37206 \dots t)

Solution steps

L^{- 1} {\frac{1}{7 s ^{2} + 37.4 s + 19990}}

Expand

\frac{1}{7 s ^{2} + 37.4 s + 19990} : \frac{1}{7} \cdot \frac{1}{( s + 2.67142 \dots ) ^{2} + 2848.57775 \dots}

= L^{- 1} {\frac{1}{7} \cdot \frac{1}{( s + 2.67142 \dots ) ^{2} + 2848.57775 \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)}

= \frac{1}{7} L^{- 1} {\frac{1}{( s + 2.67142 \dots ) ^{2} + 2848.57775 \dots}}

L^{- 1} {\frac{1}{( s + 2.67142 \dots ) ^{2} + 2848.57775 \dots}} : e^{- 2.67142 \dots t} \frac{1}{53.37206 \dots} sin (53.37206 \dots t)

= \frac{1}{7} e^{- 2.67142 \dots t} \frac{1}{53.37206 \dots} sin (53.37206 \dots t)

Simplify

\frac{1}{7} e^{- 2.67142 \dots t} \frac{1}{53.37206 \dots} sin (53.37206 \dots t) : 0.00267 \dots e^{- 2.67142 \dots t} sin (53.37206 \dots t)

= 0.00267 \dots e^{- 2.67142 \dots t} sin (53.37206 \dots t)

\frac{d}{d x} ((- 3 + x^{2})^{- 1})

\frac{( 1 + e ^{- 2 s} ) ^{2}}{s + 2}

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

\int_{- 1}^{1} x + 1 d x

What is inverse oflaplace 1/(7s^2+37.4s+19990) ?
The answer to inverse oflaplace 1/(7s^2+37.4s+19990) is 0.00267…e^{-2.67142…t}sin(53.37206…t)