What is the solution for 5/32 y^{''}+1/4 y=0 ?

The solution for 5/32 y^{''}+1/4 y=0 is y=c_{1}cos(sqrt((2^3)/5)t)+c_{2}sin(sqrt((2^3)/5)t)

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5/32 y^{''}+1/4 y=0

\frac{5}{32} y^{^{''}} + \frac{1}{4} y = 0

y = c_{1} cos (\frac{2 ^{3}}{5} t) + c_{2} sin (\frac{2 ^{3}}{5} t)

Solution steps

\frac{5}{32} y^{''} (t) + \frac{1}{4} y = 0

Solve linear ODE:

y = c_{1} cos (\frac{2 ^{3}}{5} t) + c_{2} sin (\frac{2 ^{3}}{5} t)

y = c_{1} cos (\frac{2 ^{3}}{5} t) + c_{2} sin (\frac{2 ^{3}}{5} t)

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 7

y^{^{'''}} + 13 y^{^{''}} + 43 y^{^{'}} + 31 y = 0

y^{^{''}} - 16 y = 0, y (0) = 1, y^{^{''}} (0) = 2

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

What is the solution for 5/32 y^{''}+1/4 y=0 ?
The solution for 5/32 y^{''}+1/4 y=0 is y=c_{1}cos(sqrt((2^3)/5)t)+c_{2}sin(sqrt((2^3)/5)t)