Question

I'm working on differential equations, the method of variation of parameters, but confused on how to find the complimentary solution, (Yc), of problems.

Answer 1

I noticed sometimes for instance with lets say, y'-4y'-12y=0, you can find the characteristic equation, r^2-4r-12=(r-6)(r+2)--->r1=6, r2=-2, so yc(t)=c1e^6t+c2e^-2t. But on other problems, such as y''+y=secx, the complimentary solution is yc(t)=c1cosx+c2sinx...and I am not sure why that is...