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

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

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8x^{''}+x=0

8 x^{^{''}} + x = 0

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

Solution steps

8 x^{''} (t) + x = 0

Solve linear ODE:

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

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

1.5 x^{^{''}} + 12 x^{^{'}} + 24 x = 0

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

y^{^{''}} + α^{2} y = 0

y^{^{''}} - 3 y^{^{'}} + 4 y = 0, y^{^{'}} (0) = 0, y (0) = 1

x^{^{''}} + 4 x^{^{'}} + 5 x = 0, x (0) = 1, x^{^{'}} (0) = 0

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