What is the solution for y^{''}+9y=0,y(0)=-1,y^'(0)=sqrt(3) ?

The solution for y^{''}+9y=0,y(0)=-1,y^'(0)=sqrt(3) is y=-cos(3t)+1/(sqrt(3))sin(3t)

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y^{''}+9y=0,y(0)=-1,y^'(0)=sqrt(3)

y^{^{''}} + 9 y = 0, y (0) = - 1, y^{^{'}} (0) = 3

y = - cos (3 t) + \frac{1}{3} sin (3 t)

Solution steps

y^{''} (t) + 9 y = 0

Solve linear ODE:

y = - cos (3 t) + \frac{1}{3} sin (3 t)

y = - cos (3 t) + \frac{1}{3} sin (3 t)

4 y^{^{''}} - 4 y^{^{'}} + 17 y = 0

4 y^{^{''}} - 2 y^{^{'}} + 9 y = 0

y^{^{''}} + 19 y^{^{'}} + 90 y = 0

y^{^{'''}} - 4 y^{^{''}} - 41 y^{^{'}} - 36 y = 0

y^{^{'''}} + 10 y^{^{''}} - y^{^{'}} - 3 y = 0

What is the solution for y^{''}+9y=0,y(0)=-1,y^'(0)=sqrt(3) ?
The solution for y^{''}+9y=0,y(0)=-1,y^'(0)=sqrt(3) is y=-cos(3t)+1/(sqrt(3))sin(3t)