What is the solution for 2y^{''}-4y^'+6y=0 ?

The solution for 2y^{''}-4y^'+6y=0 is y=e^t(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))

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2y^{''}-4y^'+6y=0

2 y^{^{''}} - 4 y^{^{'}} + 6 y = 0

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

Solution steps

2 y^{''} (t) - 4 y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

y^{^{'''}} + 5 y^{^{''}} + 8 y^{^{'}} + 4 y = 0

16 y^{^{''}} - 8 y^{^{'}} + 5 y = 0

4 y^{^{''}} + 36 y = 0

36 y^{^{'''}} + 60 y^{^{''}} + 25 y^{^{'}} = 0

h^{^{''}} - 2 h^{^{'}} + 10 h = 0, h (0) = 1, h^{^{'}} (0) = 10

What is the solution for 2y^{''}-4y^'+6y=0 ?
The solution for 2y^{''}-4y^'+6y=0 is y=e^t(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))