What is the solution for y^'+8(tan(8x))y=-5cos(8x) ?

The solution for y^'+8(tan(8x))y=-5cos(8x) is y=-(5x)/(sec(8x))+(c_{1})/(sec(8x))

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y^'+8(tan(8x))y=-5cos(8x)

y^{^{'}} + 8 (tan (8 x)) y = - 5 cos (8 x)

y = - \frac{5 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

Solution steps

y^{'} (x) + 8 (tan (8 x)) y = - 5 cos (8 x)

Solve linear ODE:

y = - \frac{5 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

y = - \frac{5 x}{sec ( 8 x )} + \frac{c _{1}}{sec ( 8 x )}

y = x^{2} + 7

\frac{\partial}{\partial x} (\frac{1}{cos ( x )})

(ln (2 x + 1))^{^{'}}

(2 x y) d x - (1 + y) d y = 0

What is the solution for y^'+8(tan(8x))y=-5cos(8x) ?
The solution for y^'+8(tan(8x))y=-5cos(8x) is y=-(5x)/(sec(8x))+(c_{1})/(sec(8x))