What is inverse oflaplace 1/(s^2+12s+100) ?

The answer to inverse oflaplace 1/(s^2+12s+100) is e^{-6t} 1/8 sin(8t)

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inverse oflaplace 1/(s^2+12s+100)

inverse laplace

\frac{1}{s ^{2} + 12 s + 100}

e^{- 6 t} \frac{1}{8} sin (8 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 12 s + 100}}

\frac{1}{s ^{2} + 12 s + 100} = \frac{1}{( s + 6 ) ^{2} + 64}

= L^{- 1} {\frac{1}{( s + 6 ) ^{2} + 64}}

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 + 6 ) ^{2} + 64} : a = - 6, F (s) = \frac{1}{s ^{2} + 64}

= e^{- 6 t} L^{- 1} {\frac{1}{s ^{2} + 64}}

L^{- 1} {\frac{1}{s ^{2} + 64}} : \frac{1}{8} sin (8 t)

= e^{- 6 t} \frac{1}{8} sin (8 t)

\int_{0}^{1} 1 + 12 x d x

x \to 7 lim (3)

\int (4 x + 2) e^{2 x} d x

x \to 0 lim (ln (6 x + 1))

\int \frac{x ^{3} + 16}{x ^{2} + 16} d x

What is inverse oflaplace 1/(s^2+12s+100) ?
The answer to inverse oflaplace 1/(s^2+12s+100) is e^{-6t} 1/8 sin(8t)