What is the solution for y^{''}+9y=0,y(0)=1,y^'(0)=-6 ?

The solution for y^{''}+9y=0,y(0)=1,y^'(0)=-6 is y=cos(3t)-2sin(3t)

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y^{''}+9y=0,y(0)=1,y^'(0)=-6

y^{^{''}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = - 6

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 1, y^{^{'}} (0) = 2

\frac{3 y ^{^{''}} + 2 y ^{^{'}} + y}{4} = 0

16 y^{^{''}} + y = 0

4 y^{^{''}} - 4 y^{^{'}} + 37 y = 0, y (0) = 3, y^{^{'}} (0) = 1.5

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 1

What is the solution for y^{''}+9y=0,y(0)=1,y^'(0)=-6 ?
The solution for y^{''}+9y=0,y(0)=1,y^'(0)=-6 is y=cos(3t)-2sin(3t)