Fundamental Set of Solutions

A fundamental set of solutions for an $n$ -th order linear ODE provides a basis for the solution space, from which the general solution is constructed.

Statement

Theorem3.2Existence of a fundamental set

Consider the $n$ -th order linear homogeneous ODE

$y^{(n)} + p_{n-1}(t)y^{(n-1)} + \cdots + p_1(t)y' + p_0(t)y = 0$

where $p_0, \ldots, p_{n-1}$ are continuous on an interval $I$ . Then:

The solution space is a vector space of dimension $n$ .
There exist $n$ solutions $y_1, \ldots, y_n$ that are linearly independent on $I$ .
Every solution can be written as $y = c_1 y_1 + \cdots + c_n y_n$ for unique constants $c_1, \ldots, c_n$ .
The solutions $y_1, \ldots, y_n$ are linearly independent if and only if their Wronskian $W(y_1, \ldots, y_n)(t_0) \neq 0$ for some (equivalently, every) $t_0 \in I$ .

RemarkWronskian determinant

The Wronskian of $y_1, \ldots, y_n$ is

$W(t) = \begin{vmatrix} y_1 & y_2 & \cdots & y_n \\ y_1' & y_2' & \cdots & y_n' \\ \vdots & & \ddots & \vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \cdots & y_n^{(n-1)} \end{vmatrix}.$

By Abel's formula, $W(t) = W(t_0)\exp\left(-\int_{t_0}^t p_{n-1}(s)\,ds\right)$ . Hence $W$ is either identically zero or never zero on $I$ .

Proof

Linearity: If $y_1, y_2$ are solutions and $c_1, c_2 \in \mathbb{R}$ , then $c_1y_1 + c_2y_2$ is a solution (by linearity of differentiation). So the solution set is a subspace of $C^n(I)$ .

Dimension = $n$ : For any initial conditions $(y(t_0), y'(t_0), \ldots, y^{(n-1)}(t_0)) = (a_0, a_1, \ldots, a_{n-1})$ , the existence and uniqueness theorem guarantees exactly one solution. Define $y_k$ as the solution with initial conditions given by the $k$ -th standard basis vector: $y_k^{(j)}(t_0) = \delta_{jk}$ .

Then $W(y_1, \ldots, y_n)(t_0) = \det(I_n) = 1 \neq 0$ , so $y_1, \ldots, y_n$ are linearly independent. Any solution $y$ with initial conditions $(a_0, \ldots, a_{n-1})$ equals $\sum a_k y_k$ (by uniqueness). $\blacksquare$

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Examples

ExampleThird-order constant coefficient equation

Solve $y''' - 6y'' + 11y' - 6y = 0$ .

Characteristic equation: $r^3 - 6r^2 + 11r - 6 = (r-1)(r-2)(r-3) = 0$ .

Fundamental set: $\{e^t, e^{2t}, e^{3t}\}$ . Wronskian:

$W = \begin{vmatrix} e^t & e^{2t} & e^{3t} \\ e^t & 2e^{2t} & 3e^{3t} \\ e^t & 4e^{2t} & 9e^{3t}\end{vmatrix} = 2e^{6t} \neq 0.$

General solution: $y = c_1e^t + c_2e^{2t} + c_3e^{3t}$ .

ExampleRepeated characteristic roots

For $y''' - 3y'' + 3y' - y = 0$ , the characteristic equation is $(r-1)^3 = 0$ (triple root $r = 1$ ).

Fundamental set: $\{e^t, te^t, t^2e^t\}$ . General solution: $y = (c_1 + c_2 t + c_3 t^2)e^t$ .

This illustrates the general rule: a root $r$ of multiplicity $m$ contributes $\{e^{rt}, te^{rt}, \ldots, t^{m-1}e^{rt}\}$ .

RemarkComplex roots

For an ODE with real coefficients, complex characteristic roots come in conjugate pairs $\alpha \pm i\beta$ , contributing real solutions $e^{\alpha t}\cos\beta t$ and $e^{\alpha t}\sin\beta t$ . For a pair with multiplicity $m$ , the contributions are $\{t^k e^{\alpha t}\cos\beta t, t^k e^{\alpha t}\sin\beta t\}_{k=0}^{m-1}$ .