Example: we want to solve $y'''+2y''-y'-2y = 0$. We will show, that it leads to solution of polynom $m^3 + 2 m^2 -m -2 = 0$ and the solution involves members $y= e^{mx}$.
We have learnt from linear DE of first order that the solution contains member $e^{mx}$. Without going into details, let us say that the possible solution of differential equation
$$ \begin{equation} a_n y^{(n)} + a_{n-1} y^{n-1} + \dots + a_1y' + a_0 y = 0\label{ref:DE} \end{equation} $$is
$$ \begin{equation} y = e^{mx}. \label{ref:solution} \end{equation} $$For what value of $m$ will $(\ref{ref:solution})$ be a solution of $(\ref{ref:DE})$?
$$ \begin{equation} a_n \frac{d^n}{dx^n}e^{mx} + a_{n-1}\frac{d^{n-1}}{dx^{n-1}}e^{mx} + \dots a_{1}\frac{d}{dx}e^{mx} + a_0 e^{mx} = 0. \label{ref:eq3} \end{equation} $$Since $k$-th derivative of $e^{mx} = m^k e^{mx}$, we may rewrite $(\ref{ref:eq3})$ to
$$ \begin{equation} a_n m^n e^{mx} + a_{n-1}m^{n-1}e^{mx} + \dots + a_1me^{mx} + a_0 e^{mx} = 0 \label{ref:eq4} \end{equation} $$
Let us divide $(\ref{ref:eq4})$ by $e^{mx}$ to obtain auxiliary (characteristic) equation:
$$ \begin{equation} \color{red}{{a_nm^n + a^{n-1}m^{n-1} + \dots + a_1m + a_0 = 0}}. \end{equation} $$The task diminished to an algebra task of finding roots $m$. We have to deal with the facts that
The solution consists of functions $y_1 = e^{m_1 x},\ y_2 = e^{m_2 x},\ \dots,\ y_n = e^{m_n x}$ or
Because the roots do not repeat, is is given that above terms are linearly independent.
Find the general solution of $y'''+2y''-y'-2y = 0.$
We construct the auxiliary equation:
$$ m^3 + 2m^2 -m-2 = 0 \implies m = \{-2, 1, -1 \}. $$The solution is
$$ \underline{\underline{y_c = c_1 e^{-2x} + c_2e^{x} + c_3e^{-x}}}. $$Find particular solution $y(x)$ of $y'' - 3y'+2y = 0$ for which $y_c(0) = 1$ and $y_c'(0) = 0$.
Auxiliary form is
$$ \begin{align} & m^2 - 3m +2 = 0 \implies m = \{1, 2\}. \nonumber\\ & y_c(x) = c_1e^x + c_2 e^{2x}\nonumber\\ & y_c'(x) = c_1e^x + 2 c_2 e^{2x}\nonumber\\ \end{align} $$Substitute initial values:
$$ \left. \begin{aligned} & 1 = c_1 e^0 + c_2 e^0\\ & 0 = c_1 e^0 + 2c_2 e^0 \end{aligned} \right\} \begin{aligned} &c_1=2\\ &c_2 = -1 \end{aligned} $$Substituting constants into complementaty function $y_c$ we have particular solution for given initial values:
$$ \underline{\underline{y_c = 2e^{x} - e^{2x}}}. $$Even if the roots repeat, the members of solution can not repeat. Because we expect only functions, which are linearly independent. It can be shown (reduction of order method), that if roots are $m_1 = m_2$, the solution is $e^{m_1 x} + xe^{m_2 x}$.
In general it can be shown that if auxiliary equation has a root $m=a$ which repeats $n$ times, then the general solution is
$$ y_c = (c_1 + c_2x + c_3 x^2 + \dots c_nx^n) e^{ax}. $$Find the general solution of $y(x)$ of $y^{(4)} - 3y''+2y' = 0$.
$$ \begin{aligned} &m^4 -3m^2 + 2m = 0 \implies m =\{0, 1, 1, -2\}\\ &y_c = c_1e^{0x} + c_2 e^x +c_3 x e^x + c_4 e^{-2x}\\ &\underline{\underline{y_c = c_1 + c_2 e^x +c_3 x e^x + c_4 e^{-2x}}}\\ \end{aligned} $$The imaginary root must occur in conjugate pairs: if $\alpha +\beta$ is one root, another root must be $\alpha -i\beta$.
If roots are complex $\alpha \pm i\beta$ then the solution is
$$ y_c = c_1 e^{(\alpha -i\beta)x} + c_2e^{(\alpha +i\beta)x}. $$We usually do not want to work with complex numbers. The solution above can be written also as
$$ y_c = e^{\alpha x} (c_1 \cos \beta x + c_2 \sin \beta x). $$Find the general solution of $y^{(4)} + 2y'' + y = 0$.
$$ m^4 + 2m^2 + 1 = 0 \hskip1em \text{or} \hskip1em (m^2 + 1)^2 = 0 \implies m = \{i, i, -i, -i\} $$Since complex number has a form $\alpha \pm i\beta$, we have $\beta = 1$ and the number repeats.
$$ \begin{aligned} & y_c = c_1e^{ix} + c_2^{-ix} + c_3e^{ix}x + c_4e^{-ix}x\\ & \underline{\underline{y_c = (c_1 + c_3)x e^{ix} +(c_2 + c_4)x e^{-ix} }}\\ & \text{or}\\ & y_c = e^{0}(c_1 \cos x + c_2 \sin x) + e^{0}x (c_3 \cos x + c_4 \sin x)\\ & \underline{\underline{y_c = (c_1 + c_3 x) \cos x+ (c_2 + c_4 x) \sin x}}\\ \end{aligned} $$Solve $y''+4y'+4y = 0$.
$$ \begin{aligned} &m^2 + 4m + 4 =0 \implies m = \{-2, -2\}\\ &\underline{\underline{y = c_1e^{-2x} + c_2xe^{-2x}}} \end{aligned} $$Solve $y^{(4)} - 2y''=0$.
$$ \begin{aligned} & m^4 - 2m^2 = 0\\ & (m-\sqrt{2})^2 m^2 = 0 \implies m= \{-\sqrt{2}, \sqrt{2}, 0 , 0\}\\ & y_c = c_1e^{-\sqrt{2}x} + c_2e^{\sqrt{2}x}+ c_3e^0 + c_4xe^{0} \\ & \underline{\underline{y_c = c_1e^{-\sqrt{2}x} + c_2e^{\sqrt{2}x}+ c_3 + c_4x}}\\ \end{aligned} $$Solve $y''-2y'+5y = 0$.
$$ \begin{aligned} &m^2 - 2m + 5 =0 \implies m = \{1 + 2i, 1-2i\}\hskip1em(\alpha = 1, \beta =2)\\ &y_c = c_1e^{(1+2i)x} + c_2 e^{(1-2i)x}\\ &\underline{\underline{y_c = e^x(c_1 \cos 2x + c_2\sin 2x)}} \end{aligned} $$