What is the solution for 16y^{''}+12y^'+9y=0 ?

The solution for 16y^{''}+12y^'+9y=0 is y=e^{-(3t)/8}(c_{1}cos((3^{3/2}t)/8)+c_{2}sin((3^{3/2}t)/8))

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16y^{''}+12y^'+9y=0

16 y^{^{''}} + 12 y^{^{'}} + 9 y = 0

y = e^{- \frac{3 t}{8}} (c_{1} cos (\frac{3 ^{\frac{3}{2}} t}{8}) + c_{2} sin (\frac{3 ^{\frac{3}{2}} t}{8}))

Solution steps

16 y^{''} (t) + 12 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{8}} (c_{1} cos (\frac{3 ^{\frac{3}{2}} t}{8}) + c_{2} sin (\frac{3 ^{\frac{3}{2}} t}{8}))

y = e^{- \frac{3 t}{8}} (c_{1} cos (\frac{3 ^{\frac{3}{2}} t}{8}) + c_{2} sin (\frac{3 ^{\frac{3}{2}} t}{8}))

What is the solution for 16y^{''}+12y^'+9y=0 ?
The solution for 16y^{''}+12y^'+9y=0 is y=e^{-(3t)/8}(c_{1}cos((3^{3/2}t)/8)+c_{2}sin((3^{3/2}t)/8))