What is the solution for y^{'''}-4y^{''}+5y^'-2y=0 ?

The solution for y^{'''}-4y^{''}+5y^'-2y=0 is y=c_{1}e^{2t}+c_{2}e^t+c_{3}te^t

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y^{'''}-4y^{''}+5y^'-2y=0

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = 0

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

Solution steps

y^{'''} (t) - 4 y^{''} (t) + 5 y^{'} (t) - 2 y = 0

Solve linear ODE:

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

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

2 y^{^{''''}} + 11 y^{^{'''}} + 18 y^{^{''}} + 4 y^{^{'}} - 8 y = 0

3 y^{^{''}} - 5 y^{^{'}} + 2 y = 0

u^{^{'''}} + u^{^{''}} - 2 H (t) = 0

16 y^{^{''}} - 8 y^{^{'}} + 145 y = 0

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

What is the solution for y^{'''}-4y^{''}+5y^'-2y=0 ?
The solution for y^{'''}-4y^{''}+5y^'-2y=0 is y=c_{1}e^{2t}+c_{2}e^t+c_{3}te^t