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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{x ^{3} - 3 x ^{2} + 4}{x ^{2}})

\frac{1}{( s - 5 ) ^{4}}

y^{^{'}} + tan (2 \cdot x) y = 0

\int 2 x (x^{2} - 2) d x

f (x) = (x + 2) 2 x - 3, at x = 2

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