What is inverse oflaplace (6s^2)/(s^2(s^2+16)) ?

The answer to inverse oflaplace (6s^2)/(s^2(s^2+16)) is 3/2 sin(4t)

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inverse oflaplace (6s^2)/(s^2(s^2+16))

inverse laplace

\frac{6 s ^{2}}{s ^{2} ( s ^{2} + 16 )}

\frac{3}{2} sin (4 t)

Solution steps

L^{- 1} {\frac{6 s ^{2}}{s ^{2} ( s ^{2} + 16 )}}

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)}

= 6 L^{- 1} {\frac{s ^{2}}{s ^{2} ( s ^{2} + 16 )}}

L^{- 1} {\frac{s ^{2}}{s ^{2} ( s ^{2} + 16 )}} : \frac{1}{4} sin (4 t)

= 6 \cdot \frac{1}{4} sin (4 t)

Refine

6 \frac{1}{4} sin (4 t) : \frac{3}{2} sin (4 t)

= \frac{3}{2} sin (4 t)

\frac{\partial}{\partial x} (\frac{3 x - 5}{2 x + 4})

x \to 0 + lim (\frac{14 x ^{3} + 15 x}{x})

\int - 4 sin (4 t) d t

\frac{d y}{d x} = 6 e^{x}

\frac{\partial}{\partial t} (- 2 s^{2} + t^{2})

What is inverse oflaplace (6s^2)/(s^2(s^2+16)) ?
The answer to inverse oflaplace (6s^2)/(s^2(s^2+16)) is 3/2 sin(4t)