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

The solution for y^{''}+9y= 1/((cos(3x))),y(0)=0,y^'(0)=0 is y=(xsin(3x))/3+1/9 ln(cos(3x))cos(3x)

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

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

y = \frac{x sin ( 3 x )}{3} + \frac{1}{9} ln (cos (3 x)) cos (3 x)

Solution steps

y^{''} (x) + 9 y = \frac{1}{( cos ( 3 x ) )}

Solve linear ODE:

y = \frac{x sin ( 3 x )}{3} + \frac{1}{9} ln (cos (3 x)) cos (3 x)

y = \frac{x sin ( 3 x )}{3} + \frac{1}{9} ln (cos (3 x)) cos (3 x)

y^{^{''}} - 25 y = 2

2 y^{^{''}} + y^{^{'}} = 2

y^{^{''}} - 2 y^{^{'}} - 15 y = 4

y^{^{''}} - 3 y^{^{'}} - 10 y = 54 e^{7 t}

\frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 6 y = cos (4 x)

What is the solution for y^{''}+9y= 1/((cos(3x))),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+9y= 1/((cos(3x))),y(0)=0,y^'(0)=0 is y=(xsin(3x))/3+1/9 ln(cos(3x))cos(3x)