What is the solution for y^{'''}-y^{''}+3y^'+5y=0 ?

The solution for y^{'''}-y^{''}+3y^'+5y=0 is y=c_{1}e^{-t}+e^t(c_{2}cos(2t)+c_{3}sin(2t))

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

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

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

Solution steps

y^{'''} (t) - y^{''} (t) + 3 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

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

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

64 y^{^{''}} - 48 y^{^{'}} + 9 y = 0

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

u^{^{''}} + 0.3 u^{^{'}} + 2.7 H (t) = 0

What is the solution for y^{'''}-y^{''}+3y^'+5y=0 ?
The solution for y^{'''}-y^{''}+3y^'+5y=0 is y=c_{1}e^{-t}+e^t(c_{2}cos(2t)+c_{3}sin(2t))