What is the solution for y^'+9(tan(9x))y=5cos(9x) ?

The solution for y^'+9(tan(9x))y=5cos(9x) is y=(5x)/(sec(9x))+(c_{1})/(sec(9x))

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y^'+9(tan(9x))y=5cos(9x)

y^{^{'}} + 9 (tan (9 x)) y = 5 cos (9 x)

y = \frac{5 x}{sec ( 9 x )} + \frac{c _{1}}{sec ( 9 x )}

Solution steps

y^{'} (x) + 9 (tan (9 x)) y = 5 cos (9 x)

Solve linear ODE:

y = \frac{5 x}{sec ( 9 x )} + \frac{c _{1}}{sec ( 9 x )}

y = \frac{5 x}{sec ( 9 x )} + \frac{c _{1}}{sec ( 9 x )}

\int x^{5} e^{x^{6 + 1}} d x

\frac{d ^{2}}{d x ^{2}} ((e^{x} + x^{π}) (x^{2} + x + 1))

y = x, y = - x + 2, [0, 2]

x + y = 13, x + 7 = y^{2}

f (x) = \frac{x ^{2} + 4 x}{5}

What is the solution for y^'+9(tan(9x))y=5cos(9x) ?
The solution for y^'+9(tan(9x))y=5cos(9x) is y=(5x)/(sec(9x))+(c_{1})/(sec(9x))