Complex repeated roots for ODEs

[fred2028]

I know that a 2nd order homo ordinary differential equation's solution is in the form of

\[f(x) = {C_1}{e^{{a}t}} + {C_2}t{e^{{a}t}}\]

for repeated real roots of the characteristic equation, and that the solution for a single complex root (and its conjugate) involves a cosine. I'm curious as to what the solution to a complex repeated root would look like?

 

[ZioX]

Quadratics in real numbers do not admit repeated complex roots.

 

[HallsofIvy]

If you have, say, a fourth order linear differential equation with (real) constant coefficients, then you can have repeated roots. If, say, a+ bi and a- bi are double roots, then the general solution to the differential equation is
y(x)= e^{ax}(A cos(bx)+ B sin(bx))+ xe^{ax}(C cos(bx)+ Dsin(bx))
that is, you multiply the first solutin by the independent variable just as for real roots.

(By the way, you wrote your solution as "y(x)" but then wrote the independent variable as "t". Surely, your teacher wouldn't let you get away with that on a test!)

You can have "repeated complex roots" to a second order equation if it has complex coefficients. For example, the second order, linear, differential equation with constant coefficients, y"+ 2iy'- y= 0 has characteristic equation r^2+ 2ir- 1= (r+ i)^2= 0 and so has r= -i as a double characteristic root. In that case, it would be more common to write the solution in terms of complex numbers. That is, rather than trying to use sine and cosine (which does not work if your roots are not complex conjugate), the general solution would be written y(t)= Ce^{-it}+ Dte^{-it}.