What is the solution for y^{''''}-3y^{'''}+3y^{''}-y^'=0 ?

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

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

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

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

Solution steps

y^{(4)} (t) - 3 y^{'''} (t) + 3 y^{''} (t) - y^{'} (t) = 0

Solve linear ODE:

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

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

y^{^{''''''''}} + 8 y^{^{''''}} + 16 y = 0

y^{^{''}} + y = 0, y (\frac{π}{3}) = 6, y^{^{'}} (\frac{π}{3}) = - 12

4 y^{^{''}} + 17 y^{^{'}} + 4 y = 0

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

y^{^{''}} - 8 y^{^{'}} - 33 y = 0

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