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

The solution for y^{''}+64y=0,y^'(0)=0 is y=c_{1}cos(8t)

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

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

y = c_{1} cos (8 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (8 t)

y = c_{1} cos (8 t)

x^{^{''}} + 2^{2} \cdot 3^{2} x = 0

y^{^{''}} + 4 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 3

y^{^{'''}} - y^{^{''}} - 12 y^{^{'}} - 40 y = 0, y (0) = e^{5 x}

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

x^{^{''}} + 7 x^{^{'}} + 16 x = 0

What is the solution for y^{''}+64y=0,y^'(0)=0 ?
The solution for y^{''}+64y=0,y^'(0)=0 is y=c_{1}cos(8t)