What is inverse oflaplace 1/(9s^2+1) ?

The answer to inverse oflaplace 1/(9s^2+1) is 1/3 sin(t/3)

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inverse oflaplace 1/(9s^2+1)

inverse laplace

\frac{1}{9 s ^{2} + 1}

\frac{1}{3} sin (\frac{t}{3})

Solution steps

L^{- 1} {\frac{1}{9 s ^{2} + 1}}

= L^{- 1} {\frac{1}{3} \cdot \frac{\frac{1}{3}}{s ^{2} + ( \frac{1}{3} ) ^{2}}}

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}{3} L^{- 1} {\frac{\frac{1}{3}}{s ^{2} + ( \frac{1}{3} ) ^{2}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{a}{s ^{2} + a ^{2}}} = sin (a t)

L^{- 1} {\frac{\frac{1}{3}}{s ^{2} + ( \frac{1}{3} ) ^{2}}} = sin (\frac{1}{3} t)

= \frac{1}{3} sin (\frac{1}{3} t)

= \frac{1}{3} sin (\frac{t}{3})

t \to 0 lim (e^{3 t})

y = 5 - 10 x + x^{3}, at x = - 1

\frac{x}{1 + e ^{x}}

\int \frac{5 x ^{2} - 3}{x ^{3} - x} d x

f (x) = sin (π x^{2}) - π x, at x = 1

What is inverse oflaplace 1/(9s^2+1) ?
The answer to inverse oflaplace 1/(9s^2+1) is 1/3 sin(t/3)