What is inverse oflaplace (10s)/(s^2+16) ?

The answer to inverse oflaplace (10s)/(s^2+16) is 10cos(4t)

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inverse oflaplace (10s)/(s^2+16)

inverse laplace

\frac{10 s}{s ^{2} + 16}

10 cos (4 t)

Solution steps

L^{- 1} {\frac{10 s}{s ^{2} + 16}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{10 s}{s ^{2} + 16} : F (s) = \frac{10}{s ^{2} + 16}

= \frac{d}{d t} (L^{- 1} {\frac{10}{s ^{2} + 16}}) + L^{- 1} {\frac{10}{s ^{2} + 16}} (0)

L^{- 1} {\frac{10}{s ^{2} + 16}} : \frac{5}{2} sin (4 t)

\frac{d}{d t} (\frac{5}{2} sin (4 t)) = 10 cos (4 t)

Evaluate

\frac{5}{2} sin (4 t)

at point

0 = 0

= 10 cos (4 t)

\int (sec^{2} (x) + 8) d x

\int 2 x x^{2} + 9 d x

\frac{d}{d x} ((2 x^{2} - 4 x)^{2})

\frac{3}{s s ^{2} + 8 ss + 18 s}

\frac{d}{d x} (4 x - 3 x^{2} - y^{2} - (z - 1)^{2})

What is inverse oflaplace (10s)/(s^2+16) ?
The answer to inverse oflaplace (10s)/(s^2+16) is 10cos(4t)