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}}})

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)})