What is the solution for 3y^{''}-6y^'+9y=0 ?

The solution for 3y^{''}-6y^'+9y=0 is y=e^t(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))

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3y^{''}-6y^'+9y=0

3 y^{^{''}} - 6 y^{^{'}} + 9 y = 0

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

Solution steps

3 y^{''} (t) - 6 y^{'} (t) + 9 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^{^{''}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 2

x^{''} + x = 0, x (\frac{π}{2}) = 0, x^{^{'}} (\frac{π}{2}) = 1

y^{^{'''''}} + 3 y^{^{''''}} + 4 y^{^{'''}} + 12 y^{^{''}} = 0

y^{^{'''}} - 2 y^{^{''}} - 13 y^{^{'}} - 10 y = 0

y^{^{''}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 6

What is the solution for 3y^{''}-6y^'+9y=0 ?
The solution for 3y^{''}-6y^'+9y=0 is y=e^t(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))