What is the solution for y^{''''}-4y^{'''}+6y^{''}-4y^'+y=0 ?

The solution for y^{''''}-4y^{'''}+6y^{''}-4y^'+y=0 is y=c_{1}e^t+c_{2}te^t+c_{3}t^2e^t+c_{4}t^3e^t

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y^{''''}-4y^{'''}+6y^{''}-4y^'+y=0

y^{^{''''}} - 4 y^{^{'''}} + 6 y^{^{''}} - 4 y^{^{'}} + y = 0

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

Solution steps

y^{(4)} (t) - 4 y^{'''} (t) + 6 y^{''} (t) - 4 y^{'} (t) + y = 0

Solve linear ODE:

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

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

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

x^{^{''}} + 6 x^{^{'}} + 9 x = 0

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = - 1

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

What is the solution for y^{''''}-4y^{'''}+6y^{''}-4y^'+y=0 ?
The solution for y^{''''}-4y^{'''}+6y^{''}-4y^'+y=0 is y=c_{1}e^t+c_{2}te^t+c_{3}t^2e^t+c_{4}t^3e^t