Solution of Higher-Order Linear Equations: Wronskian, Solution Space & Reduction of Order
Rigorous algebraic and analytic theory of n-th order linear differential equations: linear differential operators, the principle of superposition, linear independence, the Wronskian determinant, Abel's identity for the Wronskian, fundamental solution sets, and d'Alembert's reduction of order formula.
§5.1 The Linear Differential Operator & Superposition Principle
1. The General Linear $n$-th Order Differential Equation
The standard normalized $n$-th order linear differential equation is expressed as:
where $P_0, \dots, P_{n-1}$ and $g$ are continuous on an interval $I \subseteq \mathbb{R}$. The operator $L: C^n(I) \to C(I)$ is a linear operator, satisfying:
2. The Principle of Superposition
then any linear combination $y(x) = c_1 y_1(x) + c_2 y_2(x) + \dots + c_k y_k(x)$ is also a solution to $L[y] = 0$. Consequently, the solution space of the homogeneous equation is a vector subspace of $C^n(I)$, denoted by $\ker(L)$.
3. Dimension of the Homogeneous Solution Space
By the fundamental existence and uniqueness theorem for linear IVPs, for any fixed $x_0 \in I$, the mapping that associates each solution $y \in \ker(L)$ with its initial vector $(y(x_0), y'(x_0), \dots, y^{(n-1)}(x_0)) \in \mathbb{R}^n$ is a linear isomorphism. Thus, the solution space $\ker(L)$ has dimension exactly $n$.
§5.2 The Wronskian Determinant & Abel's Formula
1. Definition of the Wronskian Determinant
Let $y_1, y_2, \dots, y_n$ be $n$ functions that are $(n-1)$-times differentiable on an interval $I$. The Wronskian of these functions is the determinant:
2. The Wronskian Test for Linear Independence
if and only if $W(y_1, \dots, y_n)(x) \neq 0$ for all $x \in I$. Furthermore, the Wronskian is either identically zero everywhere on $I$ or never zero anywhere on $I$.
3. Proof of Abel's Identity (Abel's Formula)
Consider the second-order homogeneous equation $y'' + P(x)y' + Q(x)y = 0$, with solutions $y_1, y_2$. The Wronskian is $W(x) = y_1 y_2' - y_1' y_2$. Differentiating with respect to $x$:
Since $y_1$ and $y_2$ satisfy the ODE, $y_i'' = -P(x)y_i' - Q(x)y_i$. Substituting:
This is a first-order separable differential equation in $W(x)$: $\frac{dW}{W} = -P(x)dx$. Integrating from $x_0$ to $x$ yields Abel's formula:
Since the exponential function never vanishes, $W(x)$ is zero everywhere if $W(x_0) = 0$, and non-zero everywhere if $W(x_0) \neq 0$.
§5.3 Reduction of Order (d'Alembert's Method)
1. Theoretical Basis for Reduction of Order
If one non-trivial solution $y_1(x) \neq 0$ of the homogeneous second-order linear ODE:
is already known, a second linearly independent solution $y_2(x)$ can always be found by setting $y_2(x) = v(x) y_1(x)$, where $v(x)$ is a non-constant function to be determined.
2. Derivation of the Reduction Formula
Differentiating $y_2(x) = v y_1$:
Substituting into the ODE:
Regrouping by derivatives of $v$:
Since $y_1$ is a solution, the bracketed coefficient of $v$ vanishes identically! Setting $w = v'$ (whence $w' = v''$) reduces the equation to a first-order separable ODE for $w$:
Integrating with respect to $x$:
Integrating once more gives $v(x) = \int \frac{e^{-\int P(x) dx}}{y_1(x)^2} dx$, yielding the universal reduction of order formula:
Step-by-Step Solved Examination Problems
Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.
Compute the Wronskian of the three functions $y_1(x) = 1, y_2(x) = x, y_3(x) = x^2$. Deduce whether they form a fundamental set of solutions for a third-order linear ODE on $(-\infty, \infty)$, and construct the ODE they satisfy.
Step 1: Construct the Wronskian determinant
Step 2: Evaluate the determinant
Since the matrix is upper triangular, its determinant is simply the product of its diagonal entries:
Step 3: Linear independence and ODE construction
Since $W(x) = 2 \neq 0$ for all $x \in \mathbb{R}$, the functions are linearly independent on $(-\infty, \infty)$ and form a fundamental solution set.
Differentiating $y_3(x) = x^2$ three times gives $\frac{d^3(x^2)}{dx^3} = 0$. Likewise for $1$ and $x$. Thus, the third-order ODE is:
$W(x) = 2 \neq 0$ everywhere, confirming linear independence. The governing ODE is $y''' = 0$.
Given that $y_1(x) = x$ is a solution to the differential equation:
use the reduction of order method to find a second linearly independent solution $y_2(x)$ and write the general solution.
Step 1: Put the ODE into standard normalized form
Divide by $x^2$:
Step 2: Apply the reduction of order formula
Here $y_1(x) = x$ and $e^{-\int P(x)dx} = e^{-\int (2/x) dx} = e^{-2 \ln x} = \frac{1}{x^2}$.
Step 3: Evaluate the integral
Dropping the multiplicative constant $-1/3$ (since any non-zero multiple is linearly independent):
Step 4: Verify linear independence
Step 5: General solution
$y_2(x) = x^{-2}$. General solution: $y(x) = c_1 x + \frac{c_2}{x^2}$.
Legendre's differential equation of order $n$ is given by:
Use Abel's identity to prove that the Wronskian of any two linearly independent solutions satisfies $W(x) = \frac{C}{1 - x^2}$, and explain why $x = \pm 1$ represent singular points.
Step 1: Normalize Legendre's equation
Divide by $(1 - x^2)$:
Here $P(x) = -\frac{2x}{1 - x^2}$.
Step 2: Apply Abel's formula
By Abel's formula:
Step 3: Singular point interpretation
As $x \to \pm 1$, the denominator $1 - x^2 \to 0$, causing $W(x) \to \infty$ (unless $C = 0$). This divergence reflects the fact that $x = \pm 1$ are regular singular points of Legendre's equation, where the coefficient of $y''$ vanishes and solutions exhibit logarithmic or branch-cut singularities unless $n$ is an integer (yielding the polynomial solutions $P_n(x)$).
Proven: $W(x) = \frac{C}{1 - x^2}$. As $x \to \pm 1$, $W(x)$ diverges, identifying $x = \pm 1$ as singular points of the differential operator.