What is the solution for y^{''}+36y=0,y(0)=cos(6x) ?

The solution for y^{''}+36y=0,y(0)=cos(6x) is y=c_{1}cos(6x)+c_{2}sin(6x)

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y^{''}+36y=0,y(0)=cos(6x)

y^{^{''}} + 36 y = 0, y (0) = cos (6 x)

y = c_{1} cos (6 x) + c_{2} sin (6 x)

Solution steps

y^{''} (x) + 36 y = 0

Solve linear ODE:

y = c_{1} cos (6 x) + c_{2} sin (6 x)

y = c_{1} cos (6 x) + c_{2} sin (6 x)

5 y^{^{''}} + 30 y^{^{'}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

y^{^{''}} + 5 y^{^{'}} + 3 y = 0, y (0) = 5, y^{^{'}} (0) = 7

9 \frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} = 0

y^{^{''}} = 5 y^{^{'}} - 6 y

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

What is the solution for y^{''}+36y=0,y(0)=cos(6x) ?
The solution for y^{''}+36y=0,y(0)=cos(6x) is y=c_{1}cos(6x)+c_{2}sin(6x)