What is inverse oflaplace 4/(s^2(s^2+16)) ?

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

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inverse oflaplace 4/(s^2(s^2+16))

inverse laplace

\frac{4}{s ^{2} ( s ^{2} + 16 )}

\frac{t}{4} - \frac{1}{16} sin (4 t)

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {\frac{1}{4 s ^{2}} - \frac{1}{4 ( 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}{4 s ^{2}}} - L^{- 1} {\frac{1}{4 ( s ^{2} + 16 )}}

L^{- 1} {\frac{1}{4 s ^{2}}} : \frac{t}{4}

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

= \frac{t}{4} - \frac{1}{16} sin (4 t)

y = tan (x), at x = π

\int x^{4} (5 - x^{2}) d x

\frac{\partial}{\partial y} (- \frac{x}{( x + y ) ^{2}})

\frac{d}{d x} (\frac{x}{2 + c x})

\int \frac{1 - x ^{2}}{x ^{2}} d x

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