Question

y'+y/(sinx*cosx)=1/cosx; solve y?

Answer 1

Note that this is a linear differential equation of the form \[\frac{dy}{dx} + p(x)y = g(x)\] where \[p(x) = \frac{1}{\sin{x} \cos{x}}\]\[g(x) = \frac{1}{\cos{x}}\] Taking this into consideration, the integrating factor is \[u(x) = e^{\int p(x) \, dx} = e^{\int \frac{1}{\sin{x} \cos{x}} \, dx} = \tan{x}\] The solution is \[y(x) = \frac{\int u(x) g(x) \, dx + c}{u(x)} = \frac{\int \frac{\tan{x}}{\cos{x}} \, dx+ c}{\tan{x}} = \frac{\sec{x} + c}{\tan{x}}\]

Answer 2

thank you very much