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

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

Popular >

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)

\int_{1}^{2} e^{- x^{2}} d x

\frac{\partial}{\partial y} (x z cos (y))

y = x^{3} - 6 x^{2} + 9 x, y = x, x = 0, x = 2

\int \frac{4 e ^{x}}{e ^{x} + π} d x

\frac{d}{d x} (x^{2} sin (x) + x)

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