What is inverse oflaplace (16s+5)/(s^2+9) ?

The answer to inverse oflaplace (16s+5)/(s^2+9) is 16cos(3t)+5/3 sin(3t)

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inverse oflaplace (16s+5)/(s^2+9)

inverse laplace

\frac{16 s + 5}{s ^{2} + 9}

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

Solution steps

L^{- 1} {\frac{16 s + 5}{s ^{2} + 9}}

Expand

= L^{- 1} {\frac{16 s}{s ^{2} + 9} + \frac{5}{s ^{2} + 9}}

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)}

= 16 L^{- 1} {\frac{s}{s ^{2} + 9}} + 5 L^{- 1} {\frac{1}{s ^{2} + 9}}

L^{- 1} {\frac{s}{s ^{2} + 9}} : cos (3 t)

L^{- 1} {\frac{1}{s ^{2} + 9}} : \frac{1}{3} sin (3 t)

= 16 cos (3 t) + 5 \cdot \frac{1}{3} sin (3 t)

Refine

16 cos (3 t) + 5 \frac{1}{3} sin (3 t) : 16 cos (3 t) + \frac{5}{3} sin (3 t)

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

(f (x))^{2}

y = (2 - \frac{7}{5}) e^{3 t} + \frac{7}{5} e^{- 2 t}

3 sin (3 t)

\frac{d}{d x} ((x - 2)^{3} (6 - x)^{4})

\frac{d}{d x} (2^{x} sin (x))

What is inverse oflaplace (16s+5)/(s^2+9) ?
The answer to inverse oflaplace (16s+5)/(s^2+9) is 16cos(3t)+5/3 sin(3t)