What is inverse oflaplace ((6s-4))/(s^2-4s+20) ?

The answer to inverse oflaplace ((6s-4))/(s^2-4s+20) is 6e^{2t}cos(4t)+2e^{2t}sin(4t)

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inverse oflaplace ((6s-4))/(s^2-4s+20)

inverse laplace

\frac{( 6 s - 4 )}{s ^{2} - 4 s + 20}

6 e^{2 t} cos (4 t) + 2 e^{2 t} sin (4 t)

Solution steps

L^{- 1} {\frac{( 6 s - 4 )}{s ^{2} - 4 s + 20}}

Expand

\frac{6 s - 4}{s ^{2} - 4 s + 20} : 6 \cdot \frac{s - 2}{( s - 2 ) ^{2} + 16} + 8 \cdot \frac{1}{( s - 2 ) ^{2} + 16}

= L^{- 1} {6 \cdot \frac{s - 2}{( s - 2 ) ^{2} + 16} + 8 \cdot \frac{1}{( s - 2 ) ^{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)}

= 6 L^{- 1} {\frac{s - 2}{( s - 2 ) ^{2} + 16}} + 8 L^{- 1} {\frac{1}{( s - 2 ) ^{2} + 16}}

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

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

= 6 e^{2 t} cos (4 t) + 8 e^{2 t} \frac{1}{4} sin (4 t)

Refine

6 e^{2 t} cos (4 t) + 8 e^{2 t} \frac{1}{4} sin (4 t) : 6 e^{2 t} cos (4 t) + 2 e^{2 t} sin (4 t)

= 6 e^{2 t} cos (4 t) + 2 e^{2 t} sin (4 t)

\frac{\partial}{\partial x} (x^{\frac{2}{3}})

x \to 0 + lim (e^{x} + ln (x))

\int \frac{18}{( x + 3 ) ( x ^{2} - 9 )} d x

y^{^{''}} + 2 y^{^{'}} + y = \frac{- 4 e ^{- t}}{t ^{2} + 1}

\frac{\partial}{\partial x} (\frac{y ^{2} - x ^{2}}{x y})

What is inverse oflaplace ((6s-4))/(s^2-4s+20) ?
The answer to inverse oflaplace ((6s-4))/(s^2-4s+20) is 6e^{2t}cos(4t)+2e^{2t}sin(4t)