Solving Second Order Linear Differential Equations

Important Concepts

• Second-order linear differential equations: Often in the form y'' + p(x)y' + q(x)y = g(x), where p(x), q(x), and g(x) are continuous functions.
• Homogeneous vs. Non-homogeneous: If g(x) = 0, it's a homogeneous equation; otherwise, it's non-homogeneous.
• Constant coefficients: The equation can be simplified for constant coefficients a, b, c as ay'' + by' + cy = 0.
• Solution form: y = e^(rx).

Power House Steps

• First and second derivatives:
  • y' = re^(rx)
  • y'' = r²e^(rx)
• Substitute into the original equation:
  • Combine to form ar²e^(rx) + bre^(rx) + ce^(rx) = 0
  • Factor out e^(rx): e^(rx)(ar² + br + c) = 0
  • Simplifies to: ar² + br + c = 0

Solving the Characteristic Equation

• Standard quadratic equation: ar² + br + c = 0
• Quadratic formula: r = (-b ± sqrt(b² - 4ac)) / 2a

Three Important Cases

1. Two real roots (b² - 4ac > 0):

• r1, r2 are real and distinct
• General solution: y = C1 * e^(r1x) + C2 * e^(r2x)

2. One real root (b² - 4ac = 0):

• r is real and repeated
• General solution: y = (C1 + C2x) * e^(rx)

3. Complex roots (b² - 4ac < 0):

• Roots in the form: r = α ± βi
• General solution: y = e^(αx) * (C1*cos(βx) + C2*sin(βx))

Example Problems

Example 1: y'' - 5y' + 6y = 0

1. Set up characteristic equation: r² - 5r + 6 = 0
2. Solve by factoring: (r - 2)(r - 3) = 0
3. Roots: r = 2 and r = 3
4. General solution: y = C1 * e^(2x) + C2 * e^(3x)

Example 2: y'' - 6y' + 9y = 0

1. Set up characteristic equation: r² - 6r + 9 = 0
2. Solve by factoring: (r - 3)² = 0
3. Root: r = 3 (repeated)
4. General solution: y = (C1 + C2*x) * e^(3x)

Example 3: 9y'' + 24y' + 16y = 0

1. Set up characteristic equation: 9r² + 24r + 16 = 0
2. Factor directly: (3r + 4)² = 0
3. Root: r = -4/3 (repeated)
4. General solution: y = (C1 + C2*x) * e^(-4x/3)

Example 4: y'' + 8y' + 25y = 0

1. Set up characteristic equation: r² + 8r + 25 = 0
2. Solve using quadratic formula: r = -4 ± 3i
3. Complex roots: α = -4, β = 3
4. General solution: y = e^(-4x) * (C1*cos(3x) + C2*sin(3x))

Additional Problem: Initial Value Problem

Example: y'' + 4y = 0, y(0) = 4, y'(0) = 6

1. Characteristic equation: r² + 4 = 0, r = ±2i
2. General solution: y = C1*cos(2x) + C2*sin(2x)
3. Initial conditions:
   • y(0) = 4: C1 = 4
   • y'(0) = 6: 2C2 = 6, C2 = 3
4. Particular solution: y = 4*cos(2x) + 3*sin(2x)

Example: Boundary Value Problem

Problem: y'' - 2y' + y = 0, boundary conditions: y(0) = 3, y(1) = 7e

1. Characteristic equation: r² - 2r + 1 = 0, r = 1
2. General solution: y = (C1 + C2*x) * e^x
3. Boundary conditions:
   • y(0) = 3: C1 = 3
   • y(1) = 7e: Solving for C2 gives 4e
4. Particular solution: y = 3e^x + 4x*e^x

Summary

• Identify the type of differential equation (homogeneous or non-homogeneous).
• Use characteristic equation for constant coefficients.
• Solve using the quadratic formula and identify the type of roots.
• Write the general solution based on the type of roots.
• Apply initial or boundary conditions to find particular solutions.