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

u^{^{''}} + 0.3 u^{^{'}} + 2.7 H (t) = 0

y^{^{''}} + 5 y^{^{'}} + y = 0, y (0) = 0

y^{^{''}} - 5 y^{^{'}} - 6 y = 0, y (0) = - 1, y^{^{'}} (0) = 15

y^{^{'''}} + 48 y^{^{''}} + 768 y^{^{'}} + 4096 y = 0

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