What is inverse oflaplace 6/s+8a 1/(s^2+16) ?

The answer to inverse oflaplace 6/s+8a 1/(s^2+16) is 6\H(t)+2asin(4t)

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inverse oflaplace 6/s+8a 1/(s^2+16)

inverse laplace

\frac{6}{s} + 8 a \frac{1}{s ^{2} + 16}

6 H (t) + 2 a sin (4 t)

Solution steps

L^{- 1} {\frac{6}{s} + 8 a \frac{1}{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{6}{s}} + 8 a L^{- 1} {\frac{1}{s ^{2} + 16}}

L^{- 1} {\frac{6}{s}} : 6 H (t)

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

= 6 H (t) + 8 a \frac{1}{4} sin (4 t)

Refine

6 H (t) + 8 a \frac{1}{4} sin (4 t) : 6 H (t) + 2 a sin (4 t)

= 6 H (t) + 2 a sin (4 t)

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

\frac{d}{d x} y + y = 7

\int \frac{1}{x 9 + 4 x ^{2}} d x

(x^{2} + 1) \frac{d y}{d x} + 4 x (y - 1) = 0

\frac{d}{d x} (\frac{x - 8}{x + 8})

What is inverse oflaplace 6/s+8a 1/(s^2+16) ?
The answer to inverse oflaplace 6/s+8a 1/(s^2+16) is 6\H(t)+2asin(4t)