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....

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...

Answer 1

if you have imaginary roots, you get solutions of the form
\[c_1e^{(a+ib)x}+c_2e^{(a-ib)x}\]
\[= e^{a}(c_1e^{+ibx}+c_2e^{-ibx})\]

Answer 2

now use \( e^{ibx}= cos(bx)+isin(bx) \) and \( e^{-ibx}= cos(bx)-isin(bx) \) to rewrite the exponentials as sines and cosines.

Answer 3

Ok, thank you, would y''+y=secx be imaginary though? wouldnt it be r^2+r=0?, making it r(r+1)=0?

Answer 4

y''+y=0 gives a characteristic equation r^2+1=0

y''+y'= 0 gives the characteristic equation r^2+r=0

Your problem is the first one, where you end up with r^2 = -1, and r= +/- sqrt(-1)