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

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

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

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

y = c_{1} e^{t} + c_{2} cos (2 t) + c_{3} sin (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} cos (2 t) + c_{3} sin (2 t)

y = c_{1} e^{t} + c_{2} cos (2 t) + c_{3} sin (2 t)

9 y^{^{'''}} + 12 y^{^{''}} + 4 y^{^{'}} = 0

(D^{3} - 2 D^{2} - 5 D + 6) y = 0

16 y^{^{''}} + 88 y^{^{'}} + 121 y = 0

y^{^{'''}} - 2 y^{^{''}} - 43 y^{^{'}} - 40 y = 0

\frac{y ^{^{''}}}{y} = - 1

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