What is the solution for 16y^{''}+5y=0 ?

The solution for 16y^{''}+5y=0 is y=c_{1}cos((sqrt(5)t)/4)+c_{2}sin((sqrt(5)t)/4)

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16y^{''}+5y=0

16 y^{^{''}} + 5 y = 0

y = c_{1} cos (\frac{5 t}{4}) + c_{2} sin (\frac{5 t}{4})

Solution steps

16 y^{''} (t) + 5 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{5 t}{4}) + c_{2} sin (\frac{5 t}{4})

y = c_{1} cos (\frac{5 t}{4}) + c_{2} sin (\frac{5 t}{4})

t^{3} - t

\frac{d}{d x} (\frac{x ^{5} - 16}{2 ^{x}})

\frac{d y}{d x} + 2 x y = \frac{x}{y ^{2}}, y (0) = 0

\int 21 x^{20} d x

f (x) = \frac{2 x + 4 x ^{2}}{( 1 + 4 x ) ^{2}}

What is the solution for 16y^{''}+5y=0 ?
The solution for 16y^{''}+5y=0 is y=c_{1}cos((sqrt(5)t)/4)+c_{2}sin((sqrt(5)t)/4)