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

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

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

inverse laplace

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

δ (t) - 3 sin (3 t)

Solution steps

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

Take the partial fraction of

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

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{9}{s ^{2} + 9}} : 3 sin (3 t)

= δ (t) - 3 sin (3 t)

\int \frac{2 e ^{2 x}}{e ^{2 x} + 16 e ^{x} + 63} d x

ln (x^{6})

\int_{0}^{1} \frac{x ^{4}}{x ^{5} - 1} d x

n = 0 \sum \infty 2 (- 5)^{n}

\frac{1}{s ^{3} - 7 s}

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