What is the solution for y^'+3(tan(3x))y=-2cos(3x) ?

The solution for y^'+3(tan(3x))y=-2cos(3x) is y=-(2x)/(sec(3x))+(c_{1})/(sec(3x))

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y^'+3(tan(3x))y=-2cos(3x)

y^{^{'}} + 3 (tan (3 x)) y = - 2 cos (3 x)

y = - \frac{2 x}{sec ( 3 x )} + \frac{c _{1}}{sec ( 3 x )}

Solution steps

y^{'} (x) + 3 (tan (3 x)) y = - 2 cos (3 x)

Solve linear ODE:

y = - \frac{2 x}{sec ( 3 x )} + \frac{c _{1}}{sec ( 3 x )}

y = - \frac{2 x}{sec ( 3 x )} + \frac{c _{1}}{sec ( 3 x )}

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x \to 1 lim ((2 x - 3) \cdot (x + 1))

What is the solution for y^'+3(tan(3x))y=-2cos(3x) ?
The solution for y^'+3(tan(3x))y=-2cos(3x) is y=-(2x)/(sec(3x))+(c_{1})/(sec(3x))