What is the solution for 4/32 y^{''}+3y=0 ?

The solution for 4/32 y^{''}+3y=0 is y=c_{1}cos(2sqrt(6)t)+c_{2}sin(2sqrt(6)t)

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4/32 y^{''}+3y=0

\frac{4}{32} y^{^{''}} + 3 y = 0

y = c_{1} cos (26 t) + c_{2} sin (26 t)

Solution steps

\frac{4}{32} y^{''} (t) + 3 y = 0

Solve linear ODE:

y = c_{1} cos (26 t) + c_{2} sin (26 t)

y = c_{1} cos (26 t) + c_{2} sin (26 t)

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y^{^{'}} (\frac{π}{2}) = 12

y^{^{''}} + 6 y^{^{'}} + 2 y = 0, y (0) = 4, y^{^{'}} (0) = 0

100 y^{^{''}} - 140 y^{^{'}} + 74 y = 0, y (3) = 6, y^{^{'}} (3) = 5

y^{^{''}} - 26 y^{^{'}} + 233 y = 0

(D^{4} + 2 D^{2}) y = 0

What is the solution for 4/32 y^{''}+3y=0 ?
The solution for 4/32 y^{''}+3y=0 is y=c_{1}cos(2sqrt(6)t)+c_{2}sin(2sqrt(6)t)