What is the solution for y^{'''}-12y^{''}+48y^'-64y=0 ?

The solution for y^{'''}-12y^{''}+48y^'-64y=0 is y=c_{1}e^{4t}+c_{2}te^{4t}+c_{3}t^2e^{4t}

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y^{'''}-12y^{''}+48y^'-64y=0

y^{^{'''}} - 12 y^{^{''}} + 48 y^{^{'}} - 64 y = 0

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

Solution steps

y^{'''} (t) - 12 y^{''} (t) + 48 y^{'} (t) - 64 y = 0

Solve linear ODE:

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

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

x^{^{''}} + 4 x^{^{'}} + 20 x = 0

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

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

(D^{4} - 2 D^{3} + 2 D^{2} - 2 D + 1) y = 0

y^{^{'''}} + 5 y^{^{''}} + 7 y^{^{'}} + 3 y = 0

What is the solution for y^{'''}-12y^{''}+48y^'-64y=0 ?
The solution for y^{'''}-12y^{''}+48y^'-64y=0 is y=c_{1}e^{4t}+c_{2}te^{4t}+c_{3}t^2e^{4t}