What is the solution for y^{''''}-4y^{'''}+2y^{''}+4y^'-3y=0 ?

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

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y^{''''}-4y^{'''}+2y^{''}+4y^'-3y=0

y^{^{''''}} - 4 y^{^{'''}} + 2 y^{^{''}} + 4 y^{^{'}} - 3 y = 0

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

Solution steps

y^{(4)} (t) - 4 y^{'''} (t) + 2 y^{''} (t) + 4 y^{'} (t) - 3 y = 0

Solve linear ODE:

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

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

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

y^{^{'''}} - 6 y^{^{'}} + 11 y^{^{'}} - 6 y = 0

(D^{2} + 5 D + 6) y = 0

5000 x^{^{''}} + 105324.07 x = 0

81 y^{^{''}} - 36 y^{^{'}} + 8 y = 0

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