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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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