What is inverse oflaplace ((s+4))/(s^2+4s+29) ?

The answer to inverse oflaplace ((s+4))/(s^2+4s+29) is e^{-2t}cos(5t)+2/5 e^{-2t}sin(5t)

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inverse oflaplace ((s+4))/(s^2+4s+29)

inverse laplace

\frac{( s + 4 )}{s ^{2} + 4 s + 29}

e^{- 2 t} cos (5 t) + \frac{2}{5} e^{- 2 t} sin (5 t)

Solution steps

L^{- 1} {\frac{( s + 4 )}{s ^{2} + 4 s + 29}}

Expand

\frac{s + 4}{s ^{2} + 4 s + 29} : \frac{s + 2}{( s + 2 ) ^{2} + 25} + 2 \cdot \frac{1}{( s + 2 ) ^{2} + 25}

= L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 25} + 2 \cdot \frac{1}{( s + 2 ) ^{2} + 25}}

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

L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 25}} : e^{- 2 t} cos (5 t)

L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 25}} : e^{- 2 t} \frac{1}{5} sin (5 t)

= e^{- 2 t} cos (5 t) + 2 e^{- 2 t} \frac{1}{5} sin (5 t)

Refine

e^{- 2 t} cos (5 t) + 2 e^{- 2 t} \frac{1}{5} sin (5 t) : e^{- 2 t} cos (5 t) + \frac{2}{5} e^{- 2 t} sin (5 t)

= e^{- 2 t} cos (5 t) + \frac{2}{5} e^{- 2 t} sin (5 t)

\int 1

4 sec^{2} (x) - 6 sec^{4} (x)

y^{^{''}} + 1 y = x e^{5 x}

\frac{\partial}{\partial x} (x^{2} sin (x y + z))

2 x y^{^{'}} - 2 x y = - e^{- 2 x} y^{3}

What is inverse oflaplace ((s+4))/(s^2+4s+29) ?
The answer to inverse oflaplace ((s+4))/(s^2+4s+29) is e^{-2t}cos(5t)+2/5 e^{-2t}sin(5t)