What is the solution for 4y^{''}+16y^'+19y=0 ?

The solution for 4y^{''}+16y^'+19y=0 is y=e^{-2t}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))

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4y^{''}+16y^'+19y=0

4 y^{^{''}} + 16 y^{^{'}} + 19 y = 0

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

Solution steps

4 y^{''} (t) + 16 y^{'} (t) + 19 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (arctan (\frac{d}{x}))

\int (x - 1) e^{x} d x

y^{^{''}} + 2 y^{^{'}} = (8 t - 7) sin (3 t)

\int e^{2 t} d t

\frac{\partial}{\partial y} (x^{3} cos (y))

What is the solution for 4y^{''}+16y^'+19y=0 ?
The solution for 4y^{''}+16y^'+19y=0 is y=e^{-2t}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))