What is inverse oflaplace (5s)/(s^2(s^2+16)) ?

The answer to inverse oflaplace (5s)/(s^2(s^2+16)) is 5/16 \H(t)-5/16 cos(4t)

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inverse oflaplace (5s)/(s^2(s^2+16))

inverse laplace

\frac{5 s}{s ^{2} ( s ^{2} + 16 )}

\frac{5}{16} H (t) - \frac{5}{16} cos (4 t)

Solution steps

L^{- 1} {\frac{5 s}{s ^{2} ( s ^{2} + 16 )}}

Take the partial fraction of

\frac{5 s}{s ^{2} ( s ^{2} + 16 )} : \frac{5}{16 s} - \frac{5 s}{16 ( s ^{2} + 16 )}

= L^{- 1} {\frac{5}{16 s} - \frac{5 s}{16 ( 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{5}{16 s}} - L^{- 1} {\frac{5 s}{16 ( s ^{2} + 16 )}}

L^{- 1} {\frac{5}{16 s}} : \frac{5}{16} H (t)

L^{- 1} {\frac{5 s}{16 ( s ^{2} + 16 )}} : \frac{5}{16} cos (4 t)

= \frac{5}{16} H (t) - \frac{5}{16} cos (4 t)

n = 0 \sum \infty n sin (\frac{1}{n})

y = 3 x^{3} - 3 x, (1, 0)

x \to 2 lim (x^{2} - 4 x + 3)

\frac{\partial}{\partial x} (x 1 - x^{2})

y = 8 x, y = 0, y = x^{2} - 9

What is inverse oflaplace (5s)/(s^2(s^2+16)) ?
The answer to inverse oflaplace (5s)/(s^2(s^2+16)) is 5/16 \H(t)-5/16 cos(4t)