What is inverse oflaplace (16)/(s^2+9) ?

The answer to inverse oflaplace (16)/(s^2+9) is 16/3 sin(3t)

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inverse oflaplace (16)/(s^2+9)

inverse laplace

\frac{16}{s ^{2} + 9}

\frac{16}{3} sin (3 t)

Solution steps

L^{- 1} {\frac{16}{s ^{2} + 9}}

= L^{- 1} {\frac{16}{3} \cdot \frac{3}{s ^{2} + 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{16}{3} L^{- 1} {\frac{3}{s ^{2} + 3 ^{2}}}

Use Inverse Laplace Transform table:

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

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

= \frac{16}{3} sin (3 t)

\frac{\partial}{\partial y} (\frac{4 x}{y})

\frac{d}{d x} (- \frac{1}{2} x^{4})

x \to 4 lim (3 x^{3})

\int x^{2} cos (\frac{1}{4} x) d x

(16, 1.2), (32, 2.4)

What is inverse oflaplace (16)/(s^2+9) ?
The answer to inverse oflaplace (16)/(s^2+9) is 16/3 sin(3t)