What is the solution for y^{''}+9y=-sin(3x),y(0)=0,y^'(0)= 1/6 ?

The solution for y^{''}+9y=-sin(3x),y(0)=0,y^'(0)= 1/6 is y=(xcos(3x))/6

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y^{''}+9y=-sin(3x),y(0)=0,y^'(0)= 1/6

y^{^{''}} + 9 y = - sin (3 x), y (0) = 0, y^{^{'}} (0) = \frac{1}{6}

y = \frac{x cos ( 3 x )}{6}

Solution steps

y^{''} (x) + 9 y = - sin (3 x)

Solve linear ODE:

y = \frac{x cos ( 3 x )}{6}

y = \frac{x cos ( 3 x )}{6}

x^{3}, 0

1.1 e^{0.78 t}

\frac{d}{d x} (x^{3} (x - 9)^{2})

{e^{, at}}

\int x^{23} sin (x^{24}) d x

What is the solution for y^{''}+9y=-sin(3x),y(0)=0,y^'(0)= 1/6 ?
The solution for y^{''}+9y=-sin(3x),y(0)=0,y^'(0)= 1/6 is y=(xcos(3x))/6