What is inverse oflaplace ((2s+8))/(s^2+6s+25) ?

The answer to inverse oflaplace ((2s+8))/(s^2+6s+25) is 2e^{-3t}cos(4t)+1/2 e^{-3t}sin(4t)

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inverse oflaplace ((2s+8))/(s^2+6s+25)

inverse laplace

\frac{( 2 s + 8 )}{s ^{2} + 6 s + 25}

2 e^{- 3 t} cos (4 t) + \frac{1}{2} e^{- 3 t} sin (4 t)

Solution steps

L^{- 1} {\frac{( 2 s + 8 )}{s ^{2} + 6 s + 25}}

Expand

\frac{2 s + 8}{s ^{2} + 6 s + 25} : 2 \cdot \frac{s + 3}{( s + 3 ) ^{2} + 16} + 2 \cdot \frac{1}{( s + 3 ) ^{2} + 16}

= L^{- 1} {2 \cdot \frac{s + 3}{( s + 3 ) ^{2} + 16} + 2 \cdot \frac{1}{( s + 3 ) ^{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)}

= 2 L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 16}} + 2 L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 16}}

L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 16}} : e^{- 3 t} cos (4 t)

L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 16}} : e^{- 3 t} \frac{1}{4} sin (4 t)

= 2 e^{- 3 t} cos (4 t) + 2 e^{- 3 t} \frac{1}{4} sin (4 t)

Refine

2 e^{- 3 t} cos (4 t) + 2 e^{- 3 t} \frac{1}{4} sin (4 t) : 2 e^{- 3 t} cos (4 t) + \frac{1}{2} e^{- 3 t} sin (4 t)

= 2 e^{- 3 t} cos (4 t) + \frac{1}{2} e^{- 3 t} sin (4 t)

\int \frac{2 e ^{x} - e ^{- x}}{2 e ^{x} + e ^{- x}} d x

\int \frac{18 x}{1 - x ^{4}} d x

\frac{1}{s + 1}

\frac{\partial}{\partial x} (y e^{x} cos (yz))

\frac{d}{d x} (sin^{3} (2 x - 1))

What is inverse oflaplace ((2s+8))/(s^2+6s+25) ?
The answer to inverse oflaplace ((2s+8))/(s^2+6s+25) is 2e^{-3t}cos(4t)+1/2 e^{-3t}sin(4t)