What is the solution for 2y^{''}+8y^'+52y=0 ?

The solution for 2y^{''}+8y^'+52y=0 is y=e^{-2t}(c_{1}cos(sqrt(22)t)+c_{2}sin(sqrt(22)t))

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2y^{''}+8y^'+52y=0

2 y^{^{''}} + 8 y^{^{'}} + 52 y = 0

y = e^{- 2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

Solution steps

2 y^{''} (t) + 8 y^{'} (t) + 52 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

y = e^{- 2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (1) = e^{3 x}

4 \frac{d ^{2} y}{d x ^{2}} - 16 \frac{d y}{d x} + 16 y = 0

y^{^{''}} + 2 y^{^{'}} - 4 y = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 3

\frac{d ^{2} y}{d x ^{2}} = k^{2} y

What is the solution for 2y^{''}+8y^'+52y=0 ?
The solution for 2y^{''}+8y^'+52y=0 is y=e^{-2t}(c_{1}cos(sqrt(22)t)+c_{2}sin(sqrt(22)t))