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

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

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

inverselaplace

4 sin (4 t)

Solution steps

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

= L^{- 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)}

= 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)

= 4 sin (4 t)

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