Here I tried to find the general solution of the following linear differential equation but couldn't correctly find the answer .

Question

$$(D^2 + 2i + 1)y=0$$

here's what I've done so far, obtained auxiliary equation:

$$m^2+2im+1=0$$

Roots:

$$m_1=i(\sqrt2-1)\\m_2=-i(1+\sqrt2)$$

Which should give the general solution as:

$$y=c_1e^{i(\sqrt2-1)}+c_2e^{-i(1+\sqrt2)} $$

Which is no way similar to the expected answer!

Expected answer $$c_1e^{2x}+e^{-x}\{c_2\cos(x\sqrt3)+c_3\sin(x\sqrt3)\}$$

Answer 1

$$(D^2 + 2i + 1)y=0 \implies y''+( 2i + 1)y=0$$ You did a little mistake here $$m^2+2im+1=0$$ The characteristic polynomial should be $$m^2+(2i+1)=0$$ You can find the correct answer now

Answer 2

The expected answer has 3 basis functions, and thus should be for an ODE of order 3. Its characteristic polynomial should be $$ (λ-2)(λ+1-i\sqrt3)(λ+1+i\sqrt3)=(λ-2)(λ^2+2λ+4)=λ^3-8 $$ corresponding to the ODE $(D^3-8)y=0$. No quadratic factor looks similar to your ODE.

Answer 3

From the expected answer you could see that $\ m_1=2\ $ and $\ m_2=-1\pm \sqrt3\ i\ $.

That means $\ (m-2)(am^2+bm+c)=0 \ $ is the characteristic equation of the ODE

Then by quadratic formula $$-1\pm\sqrt3 \ i=\frac{-b\ \pm\ \sqrt{b^2-4ac}}{2a}$$

You can figure out $\ a=\frac12\ $,$\ b=1\ $,$\ c=2\ $

The characteristic equation is: $$(m-2)\left(\frac12m^2+m+2\right)=0 \implies \frac12m^3-4=0\implies m^3-8=0\\(m-2)(m^2+2m+4)=0$$

So the ODE must be, based on your expected answer:

$$y'''-8=0$$

So maybe you wrote something wrong.