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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

1 - cos (7 t)

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

x \to 0 lim ((e^{x} + 1)^{\frac{1}{x}})

x = y (y - 13), x = - y (y - 13)

\frac{d}{d x} (\frac{e ^{x}}{5 x ^{2}})

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