What is the solution for y^{'''}+4y^{''}+y^'-26y=0 ?

The solution for y^{'''}+4y^{''}+y^'-26y=0 is y=c_{1}e^{2t}+e^{-3t}(c_{2}cos(2t)+c_{3}sin(2t))

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y^{'''}+4y^{''}+y^'-26y=0

y^{^{'''}} + 4 y^{^{''}} + y^{^{'}} - 26 y = 0

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

Solution steps

y^{'''} (t) + 4 y^{''} (t) + y^{'} (t) - 26 y = 0

Solve linear ODE:

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

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

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

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

y^{^{''}} - 4 y^{^{'}} + 8 y = 0, y^{^{'}} (0) = 2

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 16

What is the solution for y^{'''}+4y^{''}+y^'-26y=0 ?
The solution for y^{'''}+4y^{''}+y^'-26y=0 is y=c_{1}e^{2t}+e^{-3t}(c_{2}cos(2t)+c_{3}sin(2t))