What is the solution for y^'+8(tan(8x))y=3cos(8x) ?

The solution for y^'+8(tan(8x))y=3cos(8x) is y=(3x)/(sec(8x))+(c_{1})/(sec(8x))

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y^'+8(tan(8x))y=3cos(8x)

y^{^{'}} + 8 (tan (8 x)) y = 3 cos (8 x)

y = \frac{3 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

Solution steps

y^{'} (x) + 8 (tan (8 x)) y = 3 cos (8 x)

Solve linear ODE:

y = \frac{3 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

y = \frac{3 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

f (x) = x^{3} - 4 x^{2} - 6 x + 8

\int \frac{1}{x} \cdot ln (x) d x

\int w d w

What is the solution for y^'+8(tan(8x))y=3cos(8x) ?
The solution for y^'+8(tan(8x))y=3cos(8x) is y=(3x)/(sec(8x))+(c_{1})/(sec(8x))