6.3 Second-Order Linear Homogeneous ODEs with Constant Coefficients

6.3 Second-Order Linear Homogeneous ODEs with Constant Coefficients

Linear differential equations with constant coefficients represent the most algebraically systematic topic within ordinary differential equations on the GRE Mathematics Subject Test. Understanding the vector space of solutions, the Wronskian determinant, Abel's formula, and root topologies of characteristic polynomials enables rapid, error-free problem solving.

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Vector Space Structure and the Principle of Superposition

Consider the general second-order linear homogeneous differential equation: $L[y] = y'' + P(x) y' + Q(x) y = 0$ where $P(x)$ and $Q(x)$ are continuous functions on an open interval $I$.

• Linear Operator: The differential operator $L = \frac{d^2}{dx^2} + P(x)\frac{d}{dx} + Q(x)$ is linear: $L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2]$.
• Principle of Superposition: If $y_1(x)$ and $y_2(x)$ are solutions to $L[y] = 0$, then any linear combination $y(x) = c_1 y_1(x) + c_2 y_2(x)$ is also a solution for arbitrary constants $c_1, c_2 \in \mathbb{R}$.
• Dimension of Solution Space: The set of all solutions to $L[y] = 0$ on $I$ forms a vector space of dimension exactly equal to the order of the equation (dimension 2 for second-order ODEs).
• Fundamental Set of Solutions: Any pair of linearly independent solutions {y_1, y_2} forms a basis for this vector space. The general solution is $y(x) = c_1 y_1(x) + c_2 y_2(x)$.

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Linear Independence, the Wronskian, and Abel's Formula

Two differentiable functions $y_1, y_2$ are linearly dependent on an interval $I$ if there exist constants $c_1, c_2$ not both zero such that $c_1 y_1(x) + c_2 y_2(x) = 0$ for all $x \in I$. Otherwise, they are linearly independent.

The Wronskian Determinant

The Wronskian of two differentiable functions $y_1$ and $y_2$ is defined by: $W(y_1, y_2)(x) = \det \begin{pmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \end{pmatrix} = y_1(x) y_2'(x) - y_1'(x) y_2(x)$

Abel's Identity

If $y_1$ and $y_2$ are solutions to $y'' + P(x)y' + Q(x)y = 0$, differentiating $W(x)$ yields: $W'(x) = (y_1 y_2'' - y_1'' y_2) = y_1(-P y_2' - Q y_2) - (-P y_1' - Q y_1)y_2 = -P(x)(y_1 y_2' - y_1' y_2) = -P(x) W(x)$ Integrating this separable first-order ODE produces Abel's Formula: $W(x) = W(x_0) \exp\left(-\int_{x_0}^x P(t)\,dt\right)$

Theorem on Fundamental Sets: Because the exponential function is strictly positive, Abel's formula implies that either $W(x) = 0$ for all $x \in I$ or $W(x) \neq 0$ for all $x \in I$. Two solutions $y_1, y_2$ of $L[y] = 0$ form a fundamental set of solutions if and only if $W(y_1, y_2)(x_0) \neq 0$ at any single point $x_0 \in I$.

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Constant-Coefficient Equations: The Characteristic Equation

For constant real coefficients $a, b, c$ with $a \neq 0$: $a y'' + b y' + c y = 0$ Substituting the exponential ansatz $y = e^{r x}$ gives $(a r^2 + b r + c)e^{r x} = 0$. Because $e^{r x} \neq 0$, $r$ must satisfy the characteristic (auxiliary) equation: $a r^2 + b r + c = 0 \implies r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The nature of the roots depends on the discriminant $\Delta = b^2 - 4ac$:

Case 1: Distinct Real Roots ($\Delta > 0$)

The roots $r_1, r_2 = \frac{-b \pm \sqrt{\Delta}}{2a}$ are real and distinct. The solutions $y_1 = e^{r_1 x}$ and $y_2 = e^{r_2 x}$ have Wronskian $W = (r_2 - r_1)e^{(r_1 + r_2)x} \neq 0$, forming a fundamental set: $y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}$

Case 2: Repeated Real Root ($\Delta = 0$)

The root $r = -\frac{b}{2a}$ has algebraic multiplicity 2. One solution is $y_1 = e^{r x}$. Reduction of order ($y_2 = v(x)e^{r x}$) reveals the second independent solution $y_2 = x e^{r x}$: $y(x) = (c_1 + c_2 x)e^{r x}$

Case 3: Complex Conjugate Roots ($\Delta < 0$)

The roots are $r = \alpha \pm i \beta$, where $\alpha = -\frac{b}{2a}$ and $\beta = \frac{\sqrt{4ac - b^2}}{2a} > 0$. Using Euler's formula $e^{(\alpha \pm i\beta)x} = e^{\alpha x}(\cos(\beta x) \pm i\sin(\beta x))$ and forming real linear combinations: $y_1(x) = e^{\alpha x} \cos(\beta x), \quad y_2(x) = e^{\alpha x} \sin(\beta x)$ The general real solution is: $y(x) = e^{\alpha x}\left(c_1 \cos(\beta x) + c_2 \sin(\beta x)\right)$

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Higher-Order Linear Constant-Coefficient ODEs

For an $n$-th order constant-coefficient homogeneous ODE: $a_n y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_1 y' + a_0 y = 0$ The characteristic polynomial is $P(r) = a_n r^n + a_{n-1} r^{n-1} + \dots + a_0 = 0$.

1. Real root $r$ of multiplicity $k$: Contributes $k$ linearly independent solutions: $e^{r x}, \; x e^{r x}, \; x^2 e^{r x}, \; \dots, \; x^{k-1} e^{r x}$
2. Complex conjugate pair $\alpha \pm i\beta$ of multiplicity $k$: Contributes $2k$ linearly independent solutions: $e^{\alpha x}\cos(\beta x), \; e^{\alpha x}\sin(\beta x), \; x e^{\alpha x}\cos(\beta x), \; x e^{\alpha x}\sin(\beta x), \; \dots, \; x^{k-1}e^{\alpha x}\cos(\beta x), \; x^{k-1}e^{\alpha x}\sin(\beta x)$

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Cauchy-Euler (Equidimensional) Equations

A second-order Cauchy-Euler equation has the standard form: $a x^2 y'' + b x y' + c y = 0, \quad x > 0$

Solution via Power Ansatz $y = x^m$

Substituting $y = x^m$, $y' = m x^{m-1}$, and $y'' = m(m - 1)x^{m-2}$ into the equation gives: $a x^2 [m(m - 1)x^{m-2}] + b x [m x^{m-1}] + c x^m = [a m(m - 1) + b m + c] x^m = 0$ Dividing by $x^m \neq 0$ yields the indicial equation: $a m^2 + (b - a)m + c = 0$

Alternative Transformation: Setting $x = e^t$ (for $x > 0$) transforms $a x^2 y'' + b x y' + c y = 0$ into a constant-coefficient equation $a \frac{d^2 y}{dt^2} + (b - a)\frac{dy}{dt} + c y = 0$ in the independent variable $t$.

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Step-by-Step Worked Problems

Problem 1: Abel's Formula for Wronskian Determination

Let $y_1(x)$ and $y_2(x)$ be two linearly independent solutions to $x^2 y'' - 2x y' + (x^2 + 2)y = 0$ on $(0, \infty)$ with $W(1) = 3$. Determine the value of $W(3)$.

Solution:

1. Put the differential equation into standard form by dividing by $x^2$: $y'' - \frac{2}{x}y' + \frac{x^2 + 2}{x^2}y = 0$
2. Identify $P(x) = -\frac{2}{x}$.
3. Apply Abel's formula: $W(x) = W(1) \exp\left(-\int_1^x P(t)\,dt\right) = 3 \exp\left(\int_1^x \frac{2}{t}\,dt\right) = 3 \exp(2 \ln x) = 3 x^2$
4. Evaluate at $x = 3$: $W(3) = 3 (3)^2 = 27$

Problem 2: Cauchy-Euler Boundary Value Problem

Solve the Cauchy-Euler equation $x^2 y'' + 3x y' + y = 0$ on $x \in [1, e]$ with boundary conditions $y(1) = 2$ and $y(e) = \frac{3}{e}$.

Solution:

1. Formulate the indicial equation with $a = 1, b = 3, c = 1$: $m(m - 1) + 3m + 1 = m^2 + 2m + 1 = (m + 1)^2 = 0$
2. The root is $m = -1$ with multiplicity 2.
3. The general solution for $x > 0$ is: $y(x) = (c_1 + c_2 \ln x)x^{-1} = \frac{c_1 + c_2 \ln x}{x}$
4. Apply boundary condition $y(1) = 2$: $y(1) = \frac{c_1 + c_2 \ln(1)}{1} = c_1 = 2$
5. Apply boundary condition $y(e) = \frac{3}{e}$: $y(e) = \frac{2 + c_2 \ln e}{e} = \frac{2 + c_2}{e} = \frac{3}{e} \implies 2 + c_2 = 3 \implies c_2 = 1$
6. The unique solution is $y(x) = \frac{2 + \ln x}{x}$.

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GRE Exam Traps & Pitfalls

Trap 1: Dropping the Leading Coefficient in Abel's Formula When applying Abel's formula to $a_2(x)y'' + a_1(x)y' + a_0(x)y = 0$, you must divide by $a_2(x)$ to obtain $P(x) = a_1(x)/a_2(x)$. Integrating $a_1(x)$ directly produces an incorrect Wronskian.

Trap 2: Vanishing Wronskian Outside the Scope of Linear ODEs A classic GRE trap tests whether $W(f_1, f_2)(x) = 0$ everywhere implies linear dependence for arbitrary differentiable functions. The answer is no: $f_1(x) = x^3$ and $f_2(x) = x^2|x|$ on $[-1, 1]$ have $W(f_1, f_2)(x) = 0$ for all $x \in [-1, 1]$, yet they are linearly independent! The equivalence $W = 0 \iff$ linear dependence holds only if $f_1, f_2$ are known solutions to a common linear ODE $y'' + Py' + Qy = 0$ with continuous coefficients.

Trap 3: Cauchy-Euler Indicial Equation Coefficient Error In $a x^2 y'' + b x y' + c y = 0$, students frequently write $a m^2 + b m + c = 0$. The correct indicial equation is $a m(m - 1) + b m + c = 0$, which simplifies to $a m^2 + (b - a)m + c = 0$. The middle coefficient is $b - a$, not $b$.