1. The problem is to solve higher-order linear homogeneous ordinary differential equations (ODEs) by finding characteristic roots. 2. The general form of such an ODE is $$a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0$$ where $a_i$ are constants. 3. To solve, we assume a solution of the form $y = e^{rt}$ and substitute into the ODE, leading to the characteristic equation: $$a_n r^n + a_{n-1} r^{n-1} + \cdots + a_1 r + a_0 = 0$$ 4. Solve this polynomial equation for roots $r$. The nature of roots determines the solution: - Distinct real roots $r_i$: general solution is $$y = c_1 e^{r_1 t} + c_2 e^{r_2 t} + \cdots + c_n e^{r_n t}$$ - Repeated roots $r$ of multiplicity $m$: solution includes terms $$e^{r t}, t e^{r t}, \ldots, t^{m-1} e^{r t}$$ - Complex roots $\alpha \pm \beta i$: solution includes terms $$e^{\alpha t} (c_1 \cos \beta t + c_2 \sin \beta t)$$ 5. Example: Solve $$y'' - 5y' + 6y = 0$$ - Characteristic equation: $$r^2 - 5r + 6 = 0$$ - Factor: $$(r - 2)(r - 3) = 0$$ - Roots: $r = 2, 3$ - General solution: $$y = c_1 e^{2t} + c_2 e^{3t}$$ This method applies similarly to higher-order equations by finding all characteristic roots and constructing the general solution accordingly.