Finding the Kernel of an ordinary differential equation

Consider the ordinary differential equations:
L1 = y" - 2y' - 3y = 0
L2 = y" - 2y' + 3y = 0

I found the formal polynomials in the variable D = d/dt to be

L1 = D^2 - 2D - 3I
L2 = D^2 - 2D + 3I

I then need to find the kernel of L1 and L2, and the general solutions to each.

I found the roots of L1 to be -1 and 3, and the gen solution to be y(t) = c1e^(-t) + c2e^(3t), but I don't know what the kernel is supposed to be. Are the roots involved in the Ker?

I found the roots of L2 to be complex, being 1 +/- i sqrt(2). I'm unsure of this answer.

Any help would be much appreciated!

1) The kernel of (D^2 - 2D - 3I) is the set of solutions to (D^2 - 2D - 3I)y = 0, which you solved.
So, the kernel equals {C₁ e^(-t) + C₂ e^(3t) : C₁, C₂ constants}.

2) Similarly, since the roots of the characteristic equation of D^2 - 2D + 3I are
r = 1 ± i√2, the kernel of (D^2 - 2D + 3I) equals {e^t (C₁ cos(t√2) + C₂ sin(t√2)) : C₁, C₂ constants}.

I hope this helps!