What is inverse oflaplace 1/(s^2+16) ?

The answer to inverse oflaplace 1/(s^2+16) is 1/4 sin(4t)

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inverse oflaplace 1/(s^2+16)

inverse laplace

\frac{1}{s ^{2} + 16}

\frac{1}{4} sin (4 t)

Solution steps

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

= L^{- 1} {\frac{1}{4} \cdot \frac{4}{s ^{2} + 4 ^{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}{4} L^{- 1} {\frac{4}{s ^{2} + 4 ^{2}}}

Use Inverse Laplace Transform table:

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

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

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

\frac{d y}{d x}, y = (sin (7 x))^{l n (x)}

\frac{s - 1}{2 s ^{2} + s + 3}

\frac{d}{d x} (5 x e^{x^{2}})

\frac{d}{d t} (2 t e^{2 t})

\frac{d}{d x} (\frac{4}{x} - \frac{1}{6 x ^{3}} + \frac{8}{x ^{5}})

What is inverse oflaplace 1/(s^2+16) ?
The answer to inverse oflaplace 1/(s^2+16) is 1/4 sin(4t)