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

The answer to inverse oflaplace (16)/(s(s^2+16)) is \H(t)-cos(4t)

Popular >

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

inverse laplace

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

H (t) - cos (4 t)

Solution steps

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

Take the partial fraction of

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

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {\frac{1}{s}} : H (t)

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

= H (t) - cos (4 t)

x \to 0 lim (\frac{2}{sin ( 3 x )})

n = 1 \sum \infty \frac{7}{2 ^{n}}

\frac{d}{d x} (\frac{9 x ^{2}}{x ^{2} + 2})

\int cos (x) cos (n x) d x

\frac{\partial}{\partial x} ((1 + \frac{2}{y}) sin (x))

What is inverse oflaplace (16)/(s(s^2+16)) ?
The answer to inverse oflaplace (16)/(s(s^2+16)) is \H(t)-cos(4t)