What is inverse oflaplace (10s^2)/(s^2+9) ?

The answer to inverse oflaplace (10s^2)/(s^2+9) is 10\delta(t)-30sin(3t)

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inverse oflaplace (10s^2)/(s^2+9)

inverse laplace

\frac{10 s ^{2}}{s ^{2} + 9}

10 δ (t) - 30 sin (3 t)

Solution steps

L^{- 1} {\frac{10 s ^{2}}{s ^{2} + 9}}

Take the partial fraction of

\frac{10 s ^{2}}{s ^{2} + 9} : 10 - \frac{90}{s ^{2} + 9}

= L^{- 1} {10 - \frac{90}{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} {10} - L^{- 1} {\frac{90}{s ^{2} + 9}}

L^{- 1} {10} : 10 δ (t)

L^{- 1} {\frac{90}{s ^{2} + 9}} : 30 sin (3 t)

= 10 δ (t) - 30 sin (3 t)

x \to 1 lim (x^{2} + 5)

y^{^{'}} = x + 4 y - 2

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

x \to 0 lim (\frac{2 ^{x} - 3 ^{x}}{x})

y = (x^{3} - 16 x)^{12}, (- 4, 0)

What is inverse oflaplace (10s^2)/(s^2+9) ?
The answer to inverse oflaplace (10s^2)/(s^2+9) is 10\delta(t)-30sin(3t)