What is the solution for 64y^{''}+16y^'+50y=0 ?

The solution for 64y^{''}+16y^'+50y=0 is y=e^{-t/8}(c_{1}cos((7t)/8)+c_{2}sin((7t)/8))

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64y^{''}+16y^'+50y=0

64 y^{^{''}} + 16 y^{^{'}} + 50 y = 0

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{7 t}{8}) + c_{2} sin (\frac{7 t}{8}))

Solution steps

64 y^{''} (t) + 16 y^{'} (t) + 50 y = 0

Solve linear ODE:

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{7 t}{8}) + c_{2} sin (\frac{7 t}{8}))

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{7 t}{8}) + c_{2} sin (\frac{7 t}{8}))

What is the solution for 64y^{''}+16y^'+50y=0 ?
The solution for 64y^{''}+16y^'+50y=0 is y=e^{-t/8}(c_{1}cos((7t)/8)+c_{2}sin((7t)/8))