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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} = - y^{^{'}}

y^{^{''}} + 2 y^{^{'}} + 8 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

y^{^{''}} + 9 y = 0, y (π) = 0, y (0) = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 0

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