What is the solution for y^{'''}-2y^{''}-40y^'-64y=0 ?

The solution for y^{'''}-2y^{''}-40y^'-64y=0 is y=c_{1}e^{-2t}+c_{2}e^{-4t}+c_{3}e^{8t}

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y^{'''}-2y^{''}-40y^'-64y=0

y^{^{'''}} - 2 y^{^{''}} - 40 y^{^{'}} - 64 y = 0

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + c_{3} e^{8 t}

Solution steps

y^{'''} (t) - 2 y^{''} (t) - 40 y^{'} (t) - 64 y = 0

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + c_{3} e^{8 t}

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + c_{3} e^{8 t}

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 29 y = 0

x^{^{''}} + 2 x^{^{'}} + 26 x = 0

\frac{d ^{2} f}{d x ^{2}} + k^{2} f (x) = 0

2 y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

x^{^{''}} + 8 x^{^{'}} = 0, x (0) = - 3, x^{^{'}} (0) = 5

What is the solution for y^{'''}-2y^{''}-40y^'-64y=0 ?
The solution for y^{'''}-2y^{''}-40y^'-64y=0 is y=c_{1}e^{-2t}+c_{2}e^{-4t}+c_{3}e^{8t}