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

The solution for y^{''}+16pi^2y=0 is y=c_{1}cos(4pit)+c_{2}sin(4pit)

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

y^{^{''}} + 16 π^{2} y = 0

y = c_{1} cos (4 π t) + c_{2} sin (4 π t)

Solution steps

y^{''} (t) + 16 π^{2} y = 0

Solve linear ODE:

y = c_{1} cos (4 π t) + c_{2} sin (4 π t)

y = c_{1} cos (4 π t) + c_{2} sin (4 π t)

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

t^{''} + 4 t (x (t, z)) = 0, t (0) = - 5, t^{^{'}} (0) = 2

x^{^{''}} + 2 x^{^{'}} + 26 x = 0, x (0) = 1, x^{^{'}} (0) = 4

y^{^{''''}} + y^{^{'''}} - 7 y^{^{''}} - y^{^{'}} + 6 y = 0

\frac{d ^{2} y}{d x ^{2}} = - k^{2} y

What is the solution for y^{''}+16pi^2y=0 ?
The solution for y^{''}+16pi^2y=0 is y=c_{1}cos(4pit)+c_{2}sin(4pit)