What is the solution for y^{''}+64y=0,y(0)= 2/3 ,y^'(0)=-4/3 ?

The solution for y^{''}+64y=0,y(0)= 2/3 ,y^'(0)=-4/3 is y= 2/3 cos(8t)-1/6 sin(8t)

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y^{''}+64y=0,y(0)= 2/3 ,y^'(0)=-4/3

y^{^{''}} + 64 y = 0, y (0) = \frac{2}{3}, y^{^{'}} (0) = - \frac{4}{3}

y = \frac{2}{3} cos (8 t) - \frac{1}{6} sin (8 t)

Solution steps

y^{''} (t) + 64 y = 0

Solve linear ODE:

y = \frac{2}{3} cos (8 t) - \frac{1}{6} sin (8 t)

y = \frac{2}{3} cos (8 t) - \frac{1}{6} sin (8 t)

(4 D^{3} + 4 D^{2} + D) y = 0

y^{^{''}} + 2 y - 3 y = 0

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = 0

x^{^{''}} + 2 x^{^{'}} + 1000 x = 0

What is the solution for y^{''}+64y=0,y(0)= 2/3 ,y^'(0)=-4/3 ?
The solution for y^{''}+64y=0,y(0)= 2/3 ,y^'(0)=-4/3 is y= 2/3 cos(8t)-1/6 sin(8t)