What is the solution for y^'+6(tan(6x))y=-2cos(6x) ?

The solution for y^'+6(tan(6x))y=-2cos(6x) is y=-(2x)/(sec(6x))+(c_{1})/(sec(6x))

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y^'+6(tan(6x))y=-2cos(6x)

y^{^{'}} + 6 (tan (6 x)) y = - 2 cos (6 x)

y = - \frac{2 x}{sec ( 6 x )} + \frac{c _{1}}{sec ( 6 x )}

Solution steps

y^{'} (x) + 6 (tan (6 x)) y = - 2 cos (6 x)

Solve linear ODE:

y = - \frac{2 x}{sec ( 6 x )} + \frac{c _{1}}{sec ( 6 x )}

y = - \frac{2 x}{sec ( 6 x )} + \frac{c _{1}}{sec ( 6 x )}

\frac{d y}{d x} = \frac{x}{sin ( y )} - \frac{9}{sin ( y )}

\int_{1}^{2 y} \frac{y}{x} d x

3 cos (2 x)

x \to 2 lim (6 x - 2)

\int_{0}^{3} x + 1 d x

What is the solution for y^'+6(tan(6x))y=-2cos(6x) ?
The solution for y^'+6(tan(6x))y=-2cos(6x) is y=-(2x)/(sec(6x))+(c_{1})/(sec(6x))