I assume this is supposed to mean
.
If you make the substitution , then this DE becomes
,
a second order linear constant coefficient homogeneous ordinary differential equation.
I'm sure you can solve for , and use this to solve for .
Yes, that is correct. The characteristic equation can be written D(D+2)(D- 6)= 0 wo D= 0, -2, and 6 are characteristic roots.
For your first problem, that's a fourth order equation and must have 4 indpendent solutions. The characteristic equation was which has D= 0 as a double root and D= 4i and D= -4i as the other roots. Because of the double root, you need both and . The general solution to the equation is .