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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int (2 x - 3) e^{x^{2} - 3 x} d x

\frac{\partial}{\partial x} (x^{5} y)

\int \frac{4 x ^{2} + 2 x + 8}{x ( x ^{2} + 2 ) ^{2}} d x

x \to + 1 lim (- \frac{3}{x - 1})

y^{^{'}} = 2 y + x e^{x}, y (0) = - 1

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