The solution of a second order linear differential equation is given as x(t)=Ce^((a+ib)t)+De^((a-ib)t).

Question

Simplify to obtain the form

Answer 1

Let u=a+ib and v=a-ib, then x=Ce^ut+De^vt.

x'=uCe^ut+vDe^vt; x"=u^2Ce^ut+v^2De^vt.

u+v=2a and uv=a^2+b^2. (x-r1)(x-r2)=x^2-(r1+r2)x+r1r2 so if we let r1=u and r2=v we have x^2-2ax+a^2+b^2=(x-u)(x-v).

x"-2ax'+(a^2+b^2)x=0 has the general solution x=Ce^ut+De^vt as can be seen by substituting for x", x' and x, a and b:

u^2Ce^ut+v^2De^vt-(u+v)(uCe^ut+vDe^vt)+uv(Ce^ut+De^vt)=0.