What is the solution for 16y^{''}+2y=0 ?

The solution for 16y^{''}+2y=0 is y=c_{1}cos(t/(2^{3/2)})+c_{2}sin(t/(2^{3/2)})

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16y^{''}+2y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = - 9 x^{2}

\frac{d}{d x} (x^{2} - 2 x^{3})

\frac{d}{d x} (\frac{x ^{5}}{6})

sin (\frac{π x}{6})

\frac{d y}{d x} = 3

What is the solution for 16y^{''}+2y=0 ?
The solution for 16y^{''}+2y=0 is y=c_{1}cos(t/(2^{3/2)})+c_{2}sin(t/(2^{3/2)})