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Sometimes the solutions to differential equations were guessed by mathematicians. There is no derivation. For the ODE $ay''+by'+cy=0$ someone guessed a solution $y(x)=e^{rx}$ then tried to figure out what value $r$ needed to be to solve the differential equation. The result of plugging in $e^{rx}$ for $y(x)$ is $e^{rx}(ar^{2}+br+c)=0$. The polynomial $ar^{2}+br+c$ is called the characteristic polynomial and solving for the roots gives the values of $r$ to solve the ODE.
One of three cases for the characteristic polynomial is two distinct real roots.
\begin{equation} ay''+by'+cy=0 \qquad y(x)=e^{rx} \end{equation} \begin{equation} e^{rx}(ar^{2}+br+c)=0 \qquad ae^{rx}(r-r_{1})(r-r_{2})=0 \end{equation}Where $r_{1}$ and $r_{2}$ are the roots of the characteristic polynomial. When $r_{1}\neq r_{2}$, the solutions to the ODE are,
\begin{equation} y_{1}(x)=e^{r_{1}x} \qquad y_{2}(x)=e^{r_{2}x} \end{equation}The general solution is,
\begin{equation} y(x)=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x} \end{equation}Find the general solution to the ODE.
\begin{equation} y'' - y = 0 \end{equation}Find the general solution to the ODE.
\begin{equation} y'' - y' - 12y = 0 \end{equation}Solve the initial value problem.
\begin{equation} \frac{1}{4}y'' - y = 0 \qquad y(0) = 1 \qquad y'(0)=2 \end{equation}Solve the initial value problem.
\begin{equation} \frac{1}{2}y'' + 2y' - y = 0 \qquad y(0) = 3 \qquad y'(0)= -4 \end{equation}Solve the IVP for a mass spring damper system.
\begin{equation} z'' + 5z' + 6z = 0 \qquad z(0) = 3 \qquad z'(0) = -2 \end{equation}