Mathematics / Pure Mathematics Differential Equations I 100% Free Open Access
Chapter 6 โ€ข Theory & Derivations

Homogeneous Linear Equations with Constant Coefficients & Cauchy-Euler Equations

Complete algebraic theory of homogeneous equations with constant coefficients: characteristic polynomials, distinct real roots, repeated root multipliers, complex conjugate roots and Euler's harmonic representation, and Cauchy-Euler (equidimensional) equations via the logarithmic transformation x = e^t.

ยง6.1 Characteristic Polynomials & Roots Classification

1. The Second-Order Constant Coefficient Equation

Consider the homogeneous linear ODE with constant real coefficients:

$$a y'' + b y' + c y = 0, \quad a \neq 0$$

Assuming a trial solution of exponential form $y(x) = e^{rx}$, we compute $y' = r e^{rx}$ and $y'' = r^2 e^{rx}$. Substituting into the ODE:

$$(a r^2 + b r + c) e^{rx} = 0$$

Since $e^{rx} \neq 0$ for all real $x$, $r$ must satisfy the characteristic (auxiliary) equation:

$$a r^2 + b r + c = 0 \implies r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

2. The Three Root Regimes

  • Case 1: Real and Distinct Roots ($\Delta = b^2 - 4ac > 0$)
    The roots $r_1 \neq r_2$ are real numbers. Two linearly independent solutions are $y_1 = e^{r_1 x}$ and $y_2 = e^{r_2 x}$. The general solution is: $$y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}$$
  • Case 2: Real and Repeated Roots ($\Delta = b^2 - 4ac = 0$)
    There is a single root of multiplicity two: $r = -b / (2a)$. One solution is $y_1 = e^{rx}$. Using reduction of order, the second independent solution is $y_2 = x e^{rx}$. The general solution is: $$y(x) = (c_1 + c_2 x) e^{rx}$$
  • Case 3: Complex Conjugate Roots ($\Delta = b^2 - 4ac < 0$)
    The roots are $r = \alpha \pm i \beta$, where $\alpha = -b/(2a)$ and $\beta = \frac{\sqrt{4ac - b^2}}{2a} > 0$. Using Euler's formula $e^{(\alpha \pm i\beta)x} = e^{\alpha x}(\cos\beta x \pm i \sin\beta x)$, the real linearly independent fundamental solutions are $y_1 = e^{\alpha x} \cos(\beta x)$ and $y_2 = e^{\alpha x} \sin(\beta x)$. The general solution is: $$y(x) = e^{\alpha x} \left[ c_1 \cos(\beta x) + c_2 \sin(\beta x) \right] = R e^{\alpha x} \cos(\beta x - \delta)$$

ยง6.2 Higher-Order Constant Coefficient Equations ($n$-th Order)

1. Generalization to $n$-th Order Equations

For an $n$-th order linear ODE $a_n y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_1 y' + a_0 y = 0$, the characteristic polynomial is:

$$P(r) = a_n r^n + a_{n-1} r^{n-1} + \dots + a_1 r + a_0 = 0$$

By the Fundamental Theorem of Algebra, $P(r)$ has exactly $n$ complex roots (counting multiplicities):

  • Each real root $r$ of multiplicity $k$ contributes $k$ linearly independent solutions: $$e^{rx}, \quad x e^{rx}, \quad x^2 e^{rx}, \quad \dots, \quad x^{k-1} e^{rx}$$
  • Each complex conjugate pair $\alpha \pm i\beta$ of multiplicity $k$ contributes $2k$ linearly independent solutions: $$\begin{aligned} e^{\alpha x} \cos\beta x, \quad x e^{\alpha x} \cos\beta x, \quad \dots, \quad x^{k-1} e^{\alpha x} \cos\beta x \\ e^{\alpha x} \sin\beta x, \quad x e^{\alpha x} \sin\beta x, \quad \dots, \quad x^{k-1} e^{\alpha x} \sin\beta x \end{aligned}$$

The sum of all these solutions multiplied by arbitrary constants $c_1, \dots, c_n$ forms the complete general solution.

ยง6.3 Cauchy-Euler (Equidimensional) Differential Equations

1. Standard Form of Cauchy-Euler Equations

A linear differential equation of the form:

$$a_n x^n \frac{d^ny}{dx^n} + a_{n-1} x^{n-1} \frac{d^{n-1}y}{dx^{n-1}} + \dots + a_1 x \frac{dy}{dx} + a_0 y = g(x)$$

where the power of $x$ matches the order of the derivative in each term, is called a Cauchy-Euler (or equidimensional) equation.

2. Second-Order Homogeneous Cauchy-Euler Equation

Consider $a x^2 y'' + b x y' + c y = 0$ for $x > 0$. We seek solutions of the form $y = x^m$. Differentiating:

$$y' = m x^{m-1}, \qquad y'' = m(m-1) x^{m-2}$$

Substituting into the ODE:

$$a x^2 [m(m-1) x^{m-2}] + b x [m x^{m-1}] + c x^m = 0 \implies [a m(m - 1) + b m + c] x^m = 0$$

This yields the indicial (auxiliary) equation:

$$a m^2 + (b - a) m + c = 0$$

3. Three Cases of Solutions for Cauchy-Euler Equations

  • Distinct Real Roots ($m_1 \neq m_2$): $y(x) = c_1 x^{m_1} + c_2 x^{m_2}$.
  • Repeated Real Root ($m_1 = m_2 = m$): The second solution is obtained by logarithmic scaling: $$y(x) = x^m (c_1 + c_2 \ln x)$$
  • Complex Conjugate Roots ($m = \alpha \pm i \beta$): Using $x^{\alpha \pm i\beta} = x^\alpha e^{\pm i \beta \ln x} = x^\alpha [\cos(\beta \ln x) \pm i \sin(\beta \ln x)]$: $$y(x) = x^\alpha \left[ c_1 \cos(\beta \ln x) + c_2 \sin(\beta \ln x) \right]$$
TIERED UNIVERSITY HONORS PROBLEMS

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.

Tier 1 โ€ข Foundational Example 6.1: Fourth-Order Constant Coefficient Initial Value Problem

Find the general solution of the fourth-order differential equation:

$$\frac{d^4y}{dx^4} - 16y = 0$$

Step 1: Write the characteristic equation

$$r^4 - 16 = 0$$



Step 2: Factor the polynomial

$$(r^2 - 4)(r^2 + 4) = 0 \implies (r - 2)(r + 2)(r^2 + 4) = 0$$



Step 3: Identify the roots
The four roots are:

$$r_1 = 2, \quad r_2 = -2, \quad r_{3,4} = \pm 2i$$



Step 4: Formulate the fundamental solutions
For real roots: $y_1 = e^{2x}, y_2 = e^{-2x}$.
For imaginary roots $\alpha = 0, \beta = 2$: $y_3 = \cos(2x), y_4 = \sin(2x)$.

Step 5: Write the general solution

$$y(x) = c_1 e^{2x} + c_2 e^{-2x} + c_3 \cos(2x) + c_4 \sin(2x)$$
Final Answer & Physical Insight

$y(x) = c_1 e^{2x} + c_2 e^{-2x} + c_3 \cos(2x) + c_4 \sin(2x)$ (or $A \cosh(2x) + B \sinh(2x) + c_3 \cos(2x) + c_4 \sin(2x)$).

Tier 2 โ€ข Analytical University Exam Example 6.2: Cauchy-Euler Equation with Complex Conjugate Roots

Solve the Cauchy-Euler boundary value problem:

$$x^2 y'' - 3x y' + 13 y = 0, \quad y(1) = 2, \quad y(e^{\pi/6}) = 0, \quad x > 0$$

Step 1: Set up the indicial equation
With $y = x^m$, the equation becomes:

$$m(m - 1) - 3m + 13 = 0 \implies m^2 - 4m + 13 = 0$$



Step 2: Solve for $m$

$$m = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = 2 \pm 3i$$

Here $\alpha = 2, \beta = 3$.

Step 3: General solution

$$y(x) = x^2 \left[ c_1 \cos(3 \ln x) + c_2 \sin(3 \ln x) \right]$$



Step 4: Apply boundary conditions
At $x = 1$ ($\ln 1 = 0$):

$$y(1) = 1^2 [c_1 \cos(0) + c_2 \sin(0)] = c_1 = 2$$

At $x = e^{\pi/6}$ ($\ln(e^{\pi/6}) = \pi/6$):

$$y(e^{\pi/6}) = (e^{\pi/6})^2 \left[ 2 \cos\left(3 \cdot \frac{\pi}{6}\right) + c_2 \sin\left(3 \cdot \frac{\pi}{6}\right) \right] = 0$$
$$e^{\pi/3} \left[ 2 \cos\left(\frac{\pi}{2}\right) + c_2 \sin\left(\frac{\pi}{2}\right) \right] = 0 \implies e^{\pi/3} [2(0) + c_2(1)] = 0 \implies c_2 = 0$$



Step 5: Final solution

$$y(x) = 2x^2 \cos(3 \ln x)$$
Final Answer & Physical Insight

$y(x) = 2x^2 \cos(3 \ln x)$.

Tier 3 โ€ข Honors Challenge Example 6.3: Critical Damping & Energy Minimization in a RLC Oscillator

An unforced RLC circuit is described by $L q'' + R q' + \frac{1}{C} q = 0$. For fixed $L = 1\text{ H}$ and $C = 0.25\text{ F}$, determine the critical damping resistance $R_{\text{crit}}$. If the circuit starts with initial charge $q(0) = Q_0$ and zero current $q'(0) = 0$, solve the critical IVP and prove that the charge never crosses zero for $t > 0$.

Step 1: Indicial equation and critical damping
Characteristic equation: $L r^2 + R r + \frac{1}{C} = 0 \implies r^2 + R r + 4 = 0$.
The discriminant is $\Delta = R^2 - 4(1)(4) = R^2 - 16$.
Critical damping occurs when $\Delta = 0 \implies R_{\\text{crit}} = \sqrt{16} = 4\\,\\Omega$.

Step 2: Repeated root solution
With $R = 4\\,\\Omega$, the repeated root is $r = -R/(2L) = -4/2 = -2$.
The general solution for the charge is:

$$q(t) = (c_1 + c_2 t) e^{-2t}$$



Step 3: Apply initial conditions
$q(0) = Q_0 \implies c_1 = Q_0$.
Compute the current (derivative):

$$q'(t) = c_2 e^{-2t} - 2(c_1 + c_2 t)e^{-2t} = (c_2 - 2c_1 - 2c_2 t)e^{-2t}$$

At $t = 0$: $q'(0) = c_2 - 2c_1 = 0 \implies c_2 = 2c_1 = 2Q_0$.
Thus:

$$q(t) = Q_0 (1 + 2t) e^{-2t}$$



Step 4: Zero-crossing proof
For $t > 0$, since $Q_0 > 0$, $1 + 2t > 1 > 0$ and $e^{-2t} > 0$. Thus $q(t) > 0$ for all $t \ge 0$. The charge monotonically decays toward zero as $t \to \infty$ without ever oscillating or crossing zero, proving the defining property of critical damping.

Final Answer & Physical Insight

$R_{\text{crit}} = 4\,\Omega$. Solution: $q(t) = Q_0 (1 + 2t) e^{-2t}$. Since $1 + 2t > 0$ for all $t > 0$, $q(t) > 0$ strictly, verifying that the charge never crosses zero.