What is the solution for y^{''}+0y^'+36y=0,y(0)=4,y^'(0)=0 ?

The solution for y^{''}+0y^'+36y=0,y(0)=4,y^'(0)=0 is y=4cos(6t)

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y^{''}+0y^'+36y=0,y(0)=4,y^'(0)=0

y^{^{''}} + 0 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 0

y = 4 cos (6 t)

Solution steps

y^{''} (t) + 0 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = 4 cos (6 t)

y = 4 cos (6 t)

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

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

y^{^{''}} - 4 y^{^{'}} - 7 y = 0

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

y^{^{''}} + 4 y^{^{'}} + cy = 0, y (0) = 1, y^{^{'}} (0) = - \frac{10}{3}

What is the solution for y^{''}+0y^'+36y=0,y(0)=4,y^'(0)=0 ?
The solution for y^{''}+0y^'+36y=0,y(0)=4,y^'(0)=0 is y=4cos(6t)