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

The solution for y^{''}+9y= 1/9 sin(x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/72 sin(x)

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y^{''}+9y= 1/9 sin(x)

y^{^{''}} + 9 y = \frac{1}{9} sin (x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{72} sin (x)

Solution steps

y^{''} (x) + 9 y = \frac{1}{9} sin (x)

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{72} sin (x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{72} sin (x)

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 7 + 7 sin (x)

y^{^{''}} + y^{^{'}} = 6 x + e^{x}

y^{^{''''}} - y = 13 t + cos (t)

y^{^{''}} + 4 y = 2 e^{3 x}

y^{^{''}} + 25 y = 3 csc^{2} (5 t)

What is the solution for y^{''}+9y= 1/9 sin(x) ?
The solution for y^{''}+9y= 1/9 sin(x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/72 sin(x)