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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} = 4 y^{^{'}} - 3 y

(D^{2} + 3 D - 10) y = 0

y^{^{''}} + 8 y^{^{'}} + 8 y = 0

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

y^{^{''}} - 2 y^{^{'}} - 15 y = 0, y (0) = 2, y^{^{'}} (0) = 6

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