What is the solution for 1/16 y^{''}+4y=0 ?

The solution for 1/16 y^{''}+4y=0 is y=c_{1}cos(8t)+c_{2}sin(8t)

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1/16 y^{''}+4y=0

\frac{1}{16} y^{^{''}} + 4 y = 0

y = c_{1} cos (8 t) + c_{2} sin (8 t)

Solution steps

\frac{1}{16} y^{''} (t) + 4 y = 0

Solve linear ODE:

y = c_{1} cos (8 t) + c_{2} sin (8 t)

y = c_{1} cos (8 t) + c_{2} sin (8 t)

y^{^{''''}} + 3 y^{^{'''}} - y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

64 y^{^{''}} - 144 y^{^{'}} + 81 y = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, (1) y = e^{2 x}

y^{^{''}} + 5 y^{^{'}} - 50 y = 0

y^{^{''}} - 440 y = 0

What is the solution for 1/16 y^{''}+4y=0 ?
The solution for 1/16 y^{''}+4y=0 is y=c_{1}cos(8t)+c_{2}sin(8t)