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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d y}{d x} = 2 x sec (y)

y^{\frac{1}{2}}

x \to 7 - lim (49 - x^{2})

f (n) = 2^{n}

f (x) = 4 sin^{5} (x)

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