What is the solution for 24/32 y^{''}+72y=0 ?

The solution for 24/32 y^{''}+72y=0 is y=c_{1}cos(4sqrt(6)t)+c_{2}sin(4sqrt(6)t)

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24/32 y^{''}+72y=0

\frac{24}{32} y^{^{''}} + 72 y = 0

y = c_{1} cos (46 t) + c_{2} sin (46 t)

Solution steps

\frac{24}{32} y^{''} (t) + 72 y = 0

Solve linear ODE:

y = c_{1} cos (46 t) + c_{2} sin (46 t)

y = c_{1} cos (46 t) + c_{2} sin (46 t)

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

6 y^{^{''''}} + 5 y^{^{'''}} + 25 y^{^{''}} + 20 y^{^{'}} + 4 y = 0

4 \frac{d ^{2} y}{d x ^{2}} - 12 \frac{d y}{d x} + 5 y = 0

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

y^{^{'''''}} + 16 y^{^{''''}} + 80 y^{^{'''}} = 0

What is the solution for 24/32 y^{''}+72y=0 ?
The solution for 24/32 y^{''}+72y=0 is y=c_{1}cos(4sqrt(6)t)+c_{2}sin(4sqrt(6)t)