What is the solution for 1/(pi^2)x^{''}+16x=0 ?

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

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1/(pi^2)x^{''}+16x=0

\frac{1}{π ^{2}} x^{^{''}} + 16 x = 0

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

Solution steps

\frac{1}{π ^{2}} x^{''} (t) + 16 x = 0

Solve linear ODE:

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

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

6 y^{^{''}} - 7 y^{^{'}} - 3 y = 0

2 y^{^{''}} - y^{^{'}} - 3 y = 0

P^{^{''}} + 2 P^{^{'}} = 0

(3 D^{3} + 5 D^{2} - 2 D) y = 0

y^{^{'''}} - y^{^{''}} = 0, y (0) = 2, y^{^{'}} (0) = 0, y^{^{''}} (0) = 1

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