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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(10 D^{3} + D^{2} - 7 D + 2) y = 0

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

y^{^{''}} + 2 y^{^{'}} + 17 y = 0, y (0) = 4, y^{^{'}} (0) = - 3

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

(4 D^{4} + 20 D^{3} + 35 D^{2} + 25 D + 6) y = 0

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