What is the solution for 9y^{''}+y=0,y(0)=-2,y^'(0)=1 ?

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

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9y^{''}+y=0,y(0)=-2,y^'(0)=1

9 y^{^{''}} + y = 0, y (0) = - 2, y^{^{'}} (0) = 1

y = - 2 cos (\frac{t}{3}) + 3 sin (\frac{t}{3})

Solution steps

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

Solve linear ODE:

y = - 2 cos (\frac{t}{3}) + 3 sin (\frac{t}{3})

y = - 2 cos (\frac{t}{3}) + 3 sin (\frac{t}{3})

u, \frac{d ^{2} u}{d x ^{2}} = a u (x)

y^{^{''}} - 7 y^{^{'}} + 8 y = 0, y (0) = 2, y^{^{'}} (0) = 7

y^{^{''}} + 3 y^{^{'}} - 18 y = 0, y (0) = 3, y^{^{'}} (0) = 18

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

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 44

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