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 )}

\int x^{2} + 7 e^{x} + C d x

y^{^{''}} + 2 y^{^{'}} = 1

y^{^{''}} - 10 y^{^{'}} + 25 y = \frac{( - 3 e ^{5 x} )}{( x ^{2} + 1 )}

\frac{d}{d x} (- 2 + \frac{7}{4 x})

12 x

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))