What is the solution for 8y^{''}+31y=0 ?

The solution for 8y^{''}+31y=0 is y=c_{1}cos(sqrt((31)/(2^3))t)+c_{2}sin(sqrt((31)/(2^3))t)

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8y^{''}+31y=0

8 y^{^{''}} + 31 y = 0

y = c_{1} cos (\frac{31}{2 ^{3}} t) + c_{2} sin (\frac{31}{2 ^{3}} t)

Solution steps

8 y^{''} (t) + 31 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{31}{2 ^{3}} t) + c_{2} sin (\frac{31}{2 ^{3}} t)

y = c_{1} cos (\frac{31}{2 ^{3}} t) + c_{2} sin (\frac{31}{2 ^{3}} t)

\frac{d y}{d x} = \frac{( 2 x + x y ^{2} )}{4 y + y x ^{2}}

\frac{d y}{d x} = \frac{6 + 3 y}{x}

2 x^{4} arctan (5 x^{3})

\int ln^{2} (2 x) d x

\frac{d}{d u} (uv)

What is the solution for 8y^{''}+31y=0 ?
The solution for 8y^{''}+31y=0 is y=c_{1}cos(sqrt((31)/(2^3))t)+c_{2}sin(sqrt((31)/(2^3))t)