What is the solution for y^'-ytan(x)=sec(x) ?

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

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y^'-ytan(x)=sec(x)

y^{^{'}} - y tan (x) = sec (x)

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

Solution steps

y^{'} (x) - y tan (x) = sec (x)

Solve linear ODE:

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

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

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