What is the solution for 2y^{''}+9y=0 ?

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

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2y^{''}+9y=0

2 y^{^{''}} + 9 y = 0

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

Solution steps

2 y^{''} (t) + 9 y = 0

Solve linear ODE:

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

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

5 f^{^{''}} (t) + 50 f^{^{'}} (t) + f (t) = 0

\frac{d ^{2} y}{d x ^{2}} = - 6 y

(2000) y^{^{''}} + 18000 y^{^{'}} + 5000 y = 0

(d^{3} - 1) y = 0

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

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