What is the solution for y^{''}+64y=0,y(0)=0,y^'(0)= 1/3 ?

The solution for y^{''}+64y=0,y(0)=0,y^'(0)= 1/3 is y= 1/24 sin(8t)

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y^{''}+64y=0,y(0)=0,y^'(0)= 1/3

y^{^{''}} + 64 y = 0, y (0) = 0, y^{^{'}} (0) = \frac{1}{3}

y = \frac{1}{24} sin (8 t)

Solution steps

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

Solve linear ODE:

y = \frac{1}{24} sin (8 t)

y = \frac{1}{24} sin (8 t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + 24 y = 0

y^{^{''}} - 9 y^{^{'}} + 5 y = 0

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

4 y^{^{''}} + 4 y^{^{'}} - 3 y = 0

What is the solution for y^{''}+64y=0,y(0)=0,y^'(0)= 1/3 ?
The solution for y^{''}+64y=0,y(0)=0,y^'(0)= 1/3 is y= 1/24 sin(8t)