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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 18 y^{^{''}} + 95 y^{^{'}} - 150 y = 0

\frac{d ^{2} y}{d x ^{2}} + 11 \frac{d y}{d x} + 28 y = 0

y^{^{''''}} + y^{^{'''}} - 5 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

y^{^{''}} - 2 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 0

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

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