What is the solution for 9y^{''}+y=0,y((3pi)/2)=6,y(0)=-8 ?

The solution for 9y^{''}+y=0,y((3pi)/2)=6,y(0)=-8 is y=-8cos(t/3)+6sin(t/3)

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9y^{''}+y=0,y((3pi)/2)=6,y(0)=-8

9 y^{^{''}} + y = 0, y (\frac{3 π}{2}) = 6, y (0) = - 8

y = - 8 cos (\frac{t}{3}) + 6 sin (\frac{t}{3})

Solution steps

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

Solve linear ODE:

y = - 8 cos (\frac{t}{3}) + 6 sin (\frac{t}{3})

y = - 8 cos (\frac{t}{3}) + 6 sin (\frac{t}{3})

x + y = 0, x = y^{2} + 3 y

\int k e^{- k x} d x

ln (x^{100})

\int_{- 2}^{2} x^{2} d x

\frac{b}{( 1 - ( \frac{y}{x} ) )}

What is the solution for 9y^{''}+y=0,y((3pi)/2)=6,y(0)=-8 ?
The solution for 9y^{''}+y=0,y((3pi)/2)=6,y(0)=-8 is y=-8cos(t/3)+6sin(t/3)