What is the solution for 3y^{''}-6y^'+30y=0 ?

The solution for 3y^{''}-6y^'+30y=0 is y=e^t(c_{1}cos(3t)+c_{2}sin(3t))

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3y^{''}-6y^'+30y=0

3 y^{^{''}} - 6 y^{^{'}} + 30 y = 0

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

Solution steps

3 y^{''} (t) - 6 y^{'} (t) + 30 y = 0

Solve linear ODE:

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

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

y^{^{'''}} = 2 y^{^{'}}

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 0

2 y^{^{''}} - 3 y^{^{'}} + 2 y = 0

64 y^{^{'''}} + 112 y^{^{''}} + 49 y^{^{'}} = 0

y^{^{''}} = 5 y, y^{0} = - 2, y^{2} = 10

What is the solution for 3y^{''}-6y^'+30y=0 ?
The solution for 3y^{''}-6y^'+30y=0 is y=e^t(c_{1}cos(3t)+c_{2}sin(3t))