9.1 Homogeneous Linear Equations with Constant Coefficients

Homogeneous linear equations with constant coefficients are a key part of solving higher-order differential equations. These equations have solutions based on the roots of their characteristic equations, which determine the form of exponential, trigonometric, or polynomial terms.

Understanding how to solve these equations is crucial for tackling more complex problems. By mastering the methods for finding general and particular solutions, you'll be better equipped to handle a wide range of differential equations in various applications.

Solution Types for Homogeneous Equations

Characteristic Equation and Root Types

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• Homogeneous linear equations with constant coefficients take the form $a_n y^{(n)} + a_{n-1} y^{(n-1)} + ... + a_1 y' + a_0 y = 0$, where $a_n, a_{n-1}, ..., a_0$ are constants and $a_n ≠ 0$
• Characteristic equation derived by substituting $y = e^{rx}$ into the homogeneous linear equation
• Roots of the characteristic equation determine the general solution form
• Real distinct roots produce exponential solutions $y = ce^{rx}$ (c is an arbitrary constant)
• Complex conjugate roots generate solutions with sine and cosine functions
• Repeated roots create solutions with terms $x^k e^{rx}$ (k is root multiplicity minus one)
• Number of linearly independent solutions equals the differential equation order

Solution Forms Based on Root Types

• Real distinct roots $r_1, r_2, ..., r_n$ yield solutions $y = c_1 e^{r_1 x} + c_2 e^{r_2 x} + ... + c_n e^{r_n x}$
• Complex conjugate roots $a ± bi$ produce solutions $y = e^{ax}(c_1 \cos(bx) + c_2 \sin(bx))$
• Repeated root $r$ with multiplicity $k$ generates solution $y = (c_1 + c_2 x + ... + c_k x^{k-1})e^{rx}$
• Examples:
  • For roots 2, -3, 5: $y = c_1e^{2x} + c_2e^{-3x} + c_3e^{5x}$
  • For roots 1 ± 2i: $y = e^x(c_1 \cos(2x) + c_2 \sin(2x))$
  • For triple root 3: $y = (c_1 + c_2x + c_3x^2)e^{3x}$

Solving Homogeneous Equations

Characteristic Equation Method

• Replace differential operator $D = d/dx$ with variable $r$ in the homogeneous linear equation
• For second-order equation $ay'' + by' + cy = 0$, characteristic equation $ar^2 + br + c = 0$
• Use quadratic formula to solve second-order characteristic equations
• Higher-order equations may require factoring or numerical methods
• Each distinct root $r_i$ contributes term $c_i e^{r_i x}$ to the general solution
• Root $r$ with multiplicity $k$ adds terms $(c_1 + c_2 x + ... + c_k x^{k-1})e^{rx}$
• Complex roots $a ± bi$ result in $e^{ax}(c_1 \cos(bx) + c_2 \sin(bx))$

Solving Process and Examples

• Steps to solve homogeneous linear equations:
  1. Form the characteristic equation
  2. Solve for the roots
  3. Write the general solution based on root types
• Example 1: Solve $y'' - 5y' + 6y = 0$
  • Characteristic equation: $r^2 - 5r + 6 = 0$
  • Roots: $r = 2$ or $r = 3$
  • General solution: $y = c_1e^{2x} + c_2e^{3x}$
• Example 2: Solve $y'' + 4y' + 4y = 0$
  • Characteristic equation: $r^2 + 4r + 4 = 0$
  • Root: $r = -2$ (repeated)
  • General solution: $y = (c_1 + c_2x)e^{-2x}$

General Solution of Homogeneous Equations

Components and Structure

• Linear combination of all linearly independent solutions from characteristic equation roots
• Number of arbitrary constants equals the differential equation order
• Represents a family of functions satisfying the homogeneous linear equation
• Verification involves substituting general solution into original equation
• Examples:
  • Third-order equation: $y = c_1e^{r_1x} + c_2e^{r_2x} + c_3e^{r_3x}$
  • Fourth-order equation with complex roots: $y = c_1e^{r_1x} + c_2e^{r_2x} + e^{ax}(c_3\cos(bx) + c_4\sin(bx))$

Verification and Properties

• Substitute general solution into original equation to confirm validity
• General solution satisfies equation for all values of arbitrary constants
• Linear independence of solution components ensures completeness
• Superposition principle applies to general solution
• Examples:
  • Verify $y = c_1e^{2x} + c_2e^{3x}$ solves $y'' - 5y' + 6y = 0$
  • Show $y = c_1\cos(2x) + c_2\sin(2x)$ solves $y'' + 4y = 0$

Particular Solution of Homogeneous Equations

Initial Conditions and Uniqueness

• Initial conditions specify function and derivative values at a point (usually x = 0)
• Number of required initial conditions equals differential equation order
• Unique solution determined by combining general solution with initial conditions
• Examples:
  • Second-order equation needs y(0) and y'(0)
  • Third-order equation requires y(0), y'(0), and y''(0)

Finding and Verifying Particular Solutions

• Steps to find particular solution:
  1. Substitute initial conditions into general solution and its derivatives
  2. Solve resulting system of linear equations for arbitrary constants
  3. Substitute constant values back into general solution
• Verify particular solution satisfies differential equation and initial conditions
• Particular solution represents specific function meeting all requirements
• Examples:
  • Solve $y'' - y = 0$ with y(0) = 1, y'(0) = 2
    • General solution: $y = c_1e^x + c_2e^{-x}$
    • Apply initial conditions to find $c_1 = 3/2$ and $c_2 = -1/2$
    • Particular solution: $y = \frac{3}{2}e^x - \frac{1}{2}e^{-x}$
  • Solve $y'' + 4y = 0$ with y(0) = 0, y'(0) = 2
    • General solution: $y = c_1\cos(2x) + c_2\sin(2x)$
    • Apply initial conditions to find $c_1 = 0$ and $c_2 = 1$
    • Particular solution: $y = \sin(2x)$