What is the solution for y^{''}+64/5 y^'+64y=0 ?

The solution for y^{''}+64/5 y^'+64y=0 is y=e^{-(32t)/5}(c_{1}cos((24t)/5)+c_{2}sin((24t)/5))

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y^{''}+64/5 y^'+64y=0

y^{^{''}} + \frac{64}{5} y^{^{'}} + 64 y = 0

y = e^{- \frac{32 t}{5}} (c_{1} cos (\frac{24 t}{5}) + c_{2} sin (\frac{24 t}{5}))

Solution steps

y^{''} (t) + \frac{64}{5} y^{'} (t) + 64 y = 0

Solve linear ODE:

y = e^{- \frac{32 t}{5}} (c_{1} cos (\frac{24 t}{5}) + c_{2} sin (\frac{24 t}{5}))

y = e^{- \frac{32 t}{5}} (c_{1} cos (\frac{24 t}{5}) + c_{2} sin (\frac{24 t}{5}))

25 y^{^{''}} - 25 y^{^{'}} - 6 y = 0

y^{^{''}} + 6 y^{^{'}} + 34 y = 0, y (0) = 8

\frac{1}{8} y^{^{''}} + y^{^{'}} + 2 y = 0

y^{^{''}} = - 6 y

y^{^{'''}} - 12 y^{^{''}} - 28 y^{^{'}} + 480 y = 0

What is the solution for y^{''}+64/5 y^'+64y=0 ?
The solution for y^{''}+64/5 y^'+64y=0 is y=e^{-(32t)/5}(c_{1}cos((24t)/5)+c_{2}sin((24t)/5))