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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 7 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = 1

755 x^{^{''}} + 2 x^{^{'}} + 10 x = 0

(D^{(6)} - 4 D^{(4)} + 4 D^{(2)}) y = 0

y^{^{''}} + 12 y^{^{'}} + 27 y = 0, y (0) = 12, y^{^{'}} (0) = - 84

y^{^{'''}} - 3 y^{^{'}} = 0

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