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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 4 y^{^{''}} + 9 y^{^{'}} = 0

y^{^{''}} + 16 y = 0, y (\frac{π}{12}) = 4, y^{^{'}} (\frac{π}{12}) = - 8

y^{^{''}} - 7 y^{^{'}} - 8 y = 0

4 y^{^{''}} + 4 y^{^{'}} + 17 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

81 y^{^{''}} + 18 y^{^{'}} + y = 0

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