What is the solution for 64y^{''}+48y^'+9y=0 ?

The solution for 64y^{''}+48y^'+9y=0 is y=c_{1}e^{-(3t)/8}+c_{2}e^{-(3t)/8}t

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64y^{''}+48y^'+9y=0

64 y^{^{''}} + 48 y^{^{'}} + 9 y = 0

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

Solution steps

64 y^{''} (t) + 48 y^{'} (t) + 9 y = 0

Solve linear ODE:

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

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

x^{^{''}} + 2 x^{^{'}} + 2 x = 0, x (0) = 1, x^{^{'}} (0) = 0

y^{^{'''}} + 5 y^{^{''}} + 4 y^{^{'}} = 0

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 11

64 y^{^{''}} - 81 y = 0

y^{^{''}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = - 4

What is the solution for 64y^{''}+48y^'+9y=0 ?
The solution for 64y^{''}+48y^'+9y=0 is y=c_{1}e^{-(3t)/8}+c_{2}e^{-(3t)/8}t