What is the solution for y^{'''}+2y^{''}+5y^'=0 ?

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

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

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

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

Solution steps

y^{'''} (t) + 2 y^{''} (t) + 5 y^{'} (t) = 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))

\frac{d ^{2} y}{d x ^{2}} + k y = 0

(D^{2} + 4)^{2} y = 0

y^{^{'''}} + 6 y^{^{''}} - 245 y^{^{'}} - 64 y = 0

- a^{2} y^{^{''}} = k y

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

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