What is the solution for y^{''}=-4y,y(0)=3,y^'(0)=-6sqrt(3) ?

The solution for y^{''}=-4y,y(0)=3,y^'(0)=-6sqrt(3) is y=3cos(2t)-3sqrt(3)sin(2t)

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y^{''}=-4y,y(0)=3,y^'(0)=-6sqrt(3)

y^{^{''}} = - 4 y, y (0) = 3, y^{^{'}} (0) = - 63

y = 3 cos (2 t) - 33 sin (2 t)

Solution steps

y^{''} (t) = - 4 y

Solve linear ODE:

y = 3 cos (2 t) - 33 sin (2 t)

y = 3 cos (2 t) - 33 sin (2 t)

y^{^{''''}} - 15 y^{^{''}} + 54 y = 0

\frac{d ^{2} y}{d x ^{2}} (1) + \frac{d y}{d x} (5) + 4 y = 0

2 y^{^{''}} - 3 y^{^{'}} - 6 y = 0

9 x^{^{''}} - 12 x^{^{'}} + 4 x = 0, x (0) = 1, x^{^{'}} (0) = - 1

y^{^{''}} + y = 0, y^{^{'}} = 5, y (0) = 9

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