Verified answer

This is of the form:

dy/dx+yp(x)=q(x)

where p(x)=sec(x)

q(x)=cos(x)

The solution is y=∫ u(x) q(x) dx / u(x) +C

where u(x) = e^∫ p(x) dx

u(x) = e^∫ sec(x) dx = e^ln | sec(x) + tan(x) |

=sec(x)+tan(x)

The solution is y=∫ (sec(x)+tan(x)) cos(x) dx / [sec(x)+tan(x)]

The solution is (sec(x)+cos(x))y =∫ (sec(x)+tan(x)) cos(x) dx

∫ (sec(x)+tan(x)) cos(x) dx = ∫ dx + ∫ sin(x) dx

= x-cos(x) +C

Thus, (secx + tanx)y = x - cosx + c

It is Linear Differential Equation!

If you learn to spell it, you might learn how to do it too!