What is the solution for 25y^{''}-80y^'+64y=0 ?

The solution for 25y^{''}-80y^'+64y=0 is y=c_{1}e^{(8t)/5}+c_{2}e^{(8t)/5}t

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25y^{''}-80y^'+64y=0

25 y^{^{''}} - 80 y^{^{'}} + 64 y = 0

y = c_{1} e^{\frac{8 t}{5}} + c_{2} e^{\frac{8 t}{5}} t

Solution steps

25 y^{''} (t) - 80 y^{'} (t) + 64 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{8 t}{5}} + c_{2} e^{\frac{8 t}{5}} t

y = c_{1} e^{\frac{8 t}{5}} + c_{2} e^{\frac{8 t}{5}} t

5 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = 0

x^{^{''}} - x^{^{'}} - 6 x = 0, x (0) = 2, x^{^{'}} (0) = - 1

x^{^{''}} + 48 x = 0

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

y^{^{'''}} + 3 y^{^{''}} - 6 y^{^{'}} - 8 y = 0

What is the solution for 25y^{''}-80y^'+64y=0 ?
The solution for 25y^{''}-80y^'+64y=0 is y=c_{1}e^{(8t)/5}+c_{2}e^{(8t)/5}t