What is the solution for 1/2 y^{''}+6y^'+26y=0 ?

The solution for 1/2 y^{''}+6y^'+26y=0 is y=e^{-6t}(c_{1}cos(4t)+c_{2}sin(4t))

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1/2 y^{''}+6y^'+26y=0

\frac{1}{2} y^{^{''}} + 6 y^{^{'}} + 26 y = 0

y = e^{- 6 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

\frac{1}{2} y^{''} (t) + 6 y^{'} (t) + 26 y = 0

Solve linear ODE:

y = e^{- 6 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{- 6 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

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

y^{^{''}} - 2 y^{^{'}} + 50 y = 0

y^{^{'''}} + 4 y^{^{''}} - 15 y^{^{'}} - 18 y = 0

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 3, y^{^{'}} (0) = 6

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 8 x = 0, x = 0, t = 0

What is the solution for 1/2 y^{''}+6y^'+26y=0 ?
The solution for 1/2 y^{''}+6y^'+26y=0 is y=e^{-6t}(c_{1}cos(4t)+c_{2}sin(4t))