What is inverse oflaplace 6/s+(32)/(s^2+16) ?

The answer to inverse oflaplace 6/s+(32)/(s^2+16) is 6\H(t)+8sin(4t)

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inverse oflaplace 6/s+(32)/(s^2+16)

inverse laplace

\frac{6}{s} + \frac{32}{s ^{2} + 16}

6 H (t) + 8 sin (4 t)

Solution steps

L^{- 1} {\frac{6}{s} + \frac{32}{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{32}{s ^{2} + 16}}

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

L^{- 1} {\frac{32}{s ^{2} + 16}} : 8 sin (4 t)

= 6 H (t) + 8 sin (4 t)

\frac{8}{( s ^{2} + 1 ) ^{2}}

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

x \to 0 lim (x (ln (x))^{2})

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

3 x^{4} + 2 e^{x}, at x = 0

What is inverse oflaplace 6/s+(32)/(s^2+16) ?
The answer to inverse oflaplace 6/s+(32)/(s^2+16) is 6\H(t)+8sin(4t)