I have these two similar ODE's: $$ u''' + 6u'' + 12u' + 8u = 0$$ and $$ u''' +6u'' + 12u' +(8+i)u = 0$$
I know that for an $n$-th order ODE there are $n$ lin-indep bases of solutions. And know how to solve for 2nd order, but how would I approach these third-order ones?
On
As you didn't not add anything to the question after saying you failed in the second equation here is basic explanation to solve Homogenous Linear ODE using characteristic polynomial:
The polynomial will yelled $n$ complex or real roots, let's say one of the roots is: $a+ib$, in this case we will consider $b\ne 0$ because if $b=0$ we have a trivial solution to our purpose.
By the characteristic polynomial way we will have to take $e$ to the power of the roots, in the complex case it is the same: $e^{a+ib}$, or more accurate, $Ae^{a+ib}$. The main problem now is that we want real valued solution, or at least the closest we can get, so we use what Euler taught us: $e^{i\theta}=\cos\theta+i\sin\theta\;\;(*)$
Using $(*)$ we get: $Ae^{a+ib}=Ae^ae^{ib}=Ae^a(\cos b+i\sin b)=e^a(C_1\cos b+C_2\sin b)$.
Now in your case, use the roots the characteristic polynomial: $(r+2)^3+i=0$
$(*)$ If you want a proof, there is a lot of ways to prove Euler's formula on the internet but the most elegant one I my opinion is: consider the function $g(x)=e^{-ix}(\cos x+i\sin x)$, from this we get $g'(x)=0$, in other words this function is a constant, and $g(0)=1$, hence $e^{-ix}(\cos x+i\sin x)=1\implies e^{ix}=\cos x+i\sin x$
Hint for first equation
$$u''' + 6u'' + 12u' + 8u = 0$$ Substitute $z=u'+2u$ then the equation becomes sipmly $$z''+4z'+4z=0$$ Polynome is $R^2+4R+4=(R+2)^2=0 \implies R=-2$ .Then $$z=c_1e^{-2x}+c_2xe^{-2x}$$ Substitute back $z=y'+2y$ $$y'+2y=c_1e^{-2x}+c_2xe^{-2x}$$ $$(ye^{2x})'=c_1+c_2x$$ integrate $$(ye^{2x})=c_1x+c_2x^2+c_3$$ $$\boxed{y(x)=e^{-2x}(c_3+c_1x+c_2x^2)}$$
I let you finish the second equation...