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

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




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

  • $\begingroup$ Welcome to the site ! Be sure that a lot of people are ready to help you but no one will do your homework. Explain what you already tried and did, show your efforts and explain where you are stuck. I am afraid that, as it is, you question will be closed very quickly because totally off-topic. $\endgroup$ – Claude Leibovici Jun 23 '18 at 17:41
  • $\begingroup$ are you sure you copied correctly this differential equation ? because WA gives this complicated solution wolframalpha.com/input/?i=y%27%27%2B(2i%2B1)y%3D0 $\endgroup$ – Isham Jun 23 '18 at 18:23

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


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.


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:


So maybe you wrote something wrong.

  • 1
    $\begingroup$ An easier way is to note that the 3 roots are proportional to the 3 cube roots of unity, i.e $m_n = 2e^{i2 n \pi/3}$. Hence $m^3 = 8$ is the characteristic polynomial $\endgroup$ – Dylan Jun 24 '18 at 11:59
  • $\begingroup$ Thanks @Dylan you're right but i wanted to do it a intuitive way for him. Because there wasn't any interpretation of the question i answered with what he gave. $\endgroup$ – MR ASSASSINS117 Jun 24 '18 at 12:43

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