What is inverse oflaplace (-3s-20)/(s^2+256) ?

The answer to inverse oflaplace (-3s-20)/(s^2+256) is -3cos(16t)-5/4 sin(16t)

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inverse oflaplace (-3s-20)/(s^2+256)

inverse laplace

\frac{- 3 s - 20}{s ^{2} + 256}

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

Solution steps

L^{- 1} {\frac{- 3 s - 20}{s ^{2} + 256}}

Expand

= L^{- 1} {- \frac{3 s}{s ^{2} + 256} - \frac{20}{s ^{2} + 256}}

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)}

= - 3 L^{- 1} {\frac{s}{s ^{2} + 256}} - 20 L^{- 1} {\frac{1}{s ^{2} + 256}}

L^{- 1} {\frac{s}{s ^{2} + 256}} : cos (16 t)

L^{- 1} {\frac{1}{s ^{2} + 256}} : \frac{1}{16} sin (16 t)

= - 3 cos (16 t) - 20 \cdot \frac{1}{16} sin (16 t)

Refine

- 3 cos (16 t) - 20 \frac{1}{16} sin (16 t) : - 3 cos (16 t) - \frac{5}{4} sin (16 t)

= - 3 cos (16 t) - \frac{5}{4} sin (16 t)

\int \frac{2 u}{u ^{2} + 1} d u

166 y^{^{''}} + 135 y^{^{'}} + 1313 y = 36 \cdot 0.7 cos (x)

y^{^{''}} + a y + b = 0

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

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

What is inverse oflaplace (-3s-20)/(s^2+256) ?
The answer to inverse oflaplace (-3s-20)/(s^2+256) is -3cos(16t)-5/4 sin(16t)