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

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

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

• 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. Jun 23, 2018 at 17:41
• 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 Jun 23, 2018 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: $$(m-2)\left(\frac12m^2+m+2\right)=0 \implies \frac12m^3-4=0\implies m^3-8=0\\(m-2)(m^2+2m+4)=0$$

$$y'''-8=0$$
• 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 Jun 24, 2018 at 11:59