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

The answer to inverse oflaplace 6/s+(16)/(s^2+16) is 6\H(t)+4sin(4t)

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inverse oflaplace 6/s+(16)/(s^2+16)

inverse laplace

\frac{6}{s} + \frac{16}{s ^{2} + 16}

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

Solution steps

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

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

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

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

\frac{d ^{2} y}{d x ^{2}} = 0

\int_{- 1}^{1} e^{x} - 1 d x

x \to - 1 + lim (x^{2} - 3)

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

\int \frac{6 x}{e ^{3 x^{2}}} d x

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