HIGHER ORDER LINEAR EQUATIONS: BACKGROUND

General Problem Form:

n^th order equation

$$P_0(t)\frac{d^ny}{dt^n} + P_1(t)\frac{d^{n-1}y}{dt^{n-1}} + \cdots + P_n(t)y = G(t).$$

If P_i's and G are continuous for $t \in I = (\alpha, \beta)$, let

$$L[y] = \frac{d^ny}{dt^n} + p_1(t)\frac{d^{n-1}y}{dt^{n-1}} + \cdots + p_n(t)y = g(t),$$

with initial conditions at $t_0 \in I$:
$y(t_0)=y_0,\ y'(t_0) = y'_0,\ \ldots,\ y^{(n-1)}(t_0) = y^{(n-1)}_0.$

Existence and Uniqueness:

if p_i's and g are continuous for $t \in I$, then there exists a unique solution that satisfies the initial conditions for all $t \in I$.

Homogeneous Equation

has g(t) = 0.

• Linear combinations: if $y_1, y_2, \ldots, y_n$ are solutions to
  L[y]=0, then so is $y = c_1y_1(t) + c_2y_2(t) + \cdots + c_ny_n(t)$.
• Define the Wronskian $W(y_1, \ldots, y_n )(t)$ by
  
  $$W({\bf y})(t) = \left\vert\begin{array}{llll} y_1(t) &y_2(t)... ...y^{(n-1)}_2(t) &\ldots& y^{(n-1)}_n(t) \end{array}\right\vert.$$
  
  
  HIGHER ORDER LINEAR EQUATIONS: BACKGROUND CONTINUED

• If y is to satisfy the initial conditions, then
  
  $$\rule{0pt}{5pt}\hspace{-6mm} \left[\begin{array}{llll} y_1(t_... ...array}{l}y_0\\ y'_0\\ \vdots \\ y^{(n-1)}_0\end{array}\right].$$
  
  
  so $W(t_0)\neq 0$ for a unique solution.
• If the p_i's and g are continuous for $t \in I$, L[y_i]=0 for $i=1, \ldots, n$, and $W(t_0)\neq 0$ for some $t_0 \in I$, then any solution y is a linear combination of $y_1, y_2, \ldots y_n$.
• If $W(y_1, y_2, \ldots, y_n )\neq 0$ then $\{y_1, y_2, \ldots, y_n \}$ is a fundamental set of linearly independent solutions.

Nonhomogeneous Equations:

if L[Y] = g then

$$y = c_1y_1(t) + c_2y_2(t) + \cdots + c_ny_n(t) + Y(t).$$

is a general solution to the nonhomogeneous equation.

HOMOGENEOUS EQUATIONS:
CONSTANT COEFFICIENTS

General Problem Form:

n^th order equation

$$L[y] = a_0\frac{d^ny}{dt^n} + a_1\frac{d^{n-1}y}{dt^{n-1}} + \cdots + a_ny = 0,$$

with initial conditions at $t_0 \in I$:
$y(t_0)=y_0,\ y'(t_0) = y'_0,\ \ldots,\ y^{(n-1)}(t_0) = y^{(n-1)}_0.$

Characteristic Equation:

try solution e^rt, then

$$L[e^{rt}] = e^{rt}(a_0r^n + a_1r^{n-1} + \cdots + a_n) = 0,$$

so we need to solve the characteristic equation

$$Z(r) = a_0r^n + a_1r^{n-1} + \cdots + a_n = 0,$$

by factoring the characteristic polynomial

$$Z(r) = a_0(r-r_1)(r-r_2) \cdots (r-r_n) = 0.$$

Solution Types

• All roots real and distinct:
  
  $$y(t) = c_1e^{r_1t} + c_2e^{r_2t} + \cdots + c_ne^{r_nt}.$$
  
  

• Complex roots: appear in (conjugate) pairs. For each pair $\lambda \pm \mu i$, there are two terms in the solution:
  
  $$e^{\lambda t}\cos(\mu t) \mbox{\ and\ } e^{\lambda t}\sin(\mu t).$$
  
  

• Repeated roots: if a root r is repeated s times, there are s terms for root r in the solution:
  
  $$e^{rt},\ te^{rt},\ \ldots,\ t^{s-1}e^{rt}.$$