What is the solution for 9y^{''}+4y=0 ?

The solution for 9y^{''}+4y=0 is y=c_{1}cos((2t)/3)+c_{2}sin((2t)/3)

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9y^{''}+4y=0

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

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

Solution steps

9 y^{''} (t) + 4 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})

y^{^{''}} + 4 y = x^{2} + 3 e^{x}, y (0) = 0, y^{^{'}} (0) = 2

2 - 4 x

\int 9 sin (x) d x

\int_{0}^{2} 2 π x (x^{2}) d x

\frac{d}{d x} (e^{x \cdot c o s (x)})

What is the solution for 9y^{''}+4y=0 ?
The solution for 9y^{''}+4y=0 is y=c_{1}cos((2t)/3)+c_{2}sin((2t)/3)