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

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

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

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

y = - 6 cos (3 t) - 3 sin (3 t)

Solution steps

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

Solve linear ODE:

y = - 6 cos (3 t) - 3 sin (3 t)

y = - 6 cos (3 t) - 3 sin (3 t)

\frac{d}{d x} (\frac{- 8 x}{( x ^{2} - 4 ) ^{2}})

\frac{d}{d θ} (4 sin (θ))

\frac{d}{d x} (16 x^{3} - 96 x^{2} + 178 x - 95)

\frac{d}{d x} (- \frac{6}{1 - 6 x})

y^{^{'}} \frac{1}{x} y = 0

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