Slide 1

  • Topic: Integral Calculus - Linear ODE problem
  • Introduction to Linear Ordinary Differential Equations problem

Slide 2

  • What is an Ordinary Differential Equation (ODE)?
  • Definition of a Linear ODE
  • General form of a Linear ODE
  • Linear ODE examples:
    • dy/dx + y = x
    • 2y’’ + 3y’ + 5y = 0

Slide 3

  • Homogeneous vs. Non-Homogeneous Linear ODEs
  • Homogeneous Linear ODE:
    • A linear ODE in the form of ay’’ + by’ + cy = 0, where a, b, and c are constants
    • Example: 2y’’ + 3y’ + 5y = 0
  • Non-Homogeneous Linear ODE:
    • A linear ODE in the form of ay’’ + by’ + cy = f(x), where f(x) is a non-zero function
    • Example: 2y’’ + 3y’ + 5y = x

Slide 4

  • Order of a Linear ODE:
    • The highest derivative present in the equation determines the order
    • Example: 2y’’ + 3y’ + 5y = x is a second-order ODE
  • Linear ODE with constant coefficients:
    • Coefficients a, b, and c are constants
    • Example: 2y’’ + 3y’ + 5y = 0

Slide 5

  • Solving Linear ODE problems using the method of Integrating Factors
  • Steps for solving a Linear ODE using Integrating Factors:
    1. Identify the coefficient of the highest order derivative
    2. Let this coefficient be p(x)
    3. Find the integrating factor: IF = exp(∫p(x)dx)

Slide 6

  • Solving Linear ODE problems using the method of Integrating Factors (contd.)
  • Steps for solving a Linear ODE using Integrating Factors (contd.): 4. Multiply the entire equation by the integrating factor 5. Simplify and rearrange the equation 6. Integrate both sides with respect to x
  • Example: Solve the ODE 2y’’ + 3y’ + 5y = x

Slide 7

  • Solution for the ODE 2y’’ + 3y’ + 5y = x:
    1. Identify p(x): p(x) = 3
    2. Find the integrating factor: IF = exp(∫3dx) = exp(3x)
    3. Multiply the equation by the IF:
      • 2(exp(3x))y’’ + 3(exp(3x))y’ + 5(exp(3x))y = x(exp(3x))

Slide 8

  • Solution for the ODE 2y’’ + 3y’ + 5y = x (contd.): 4. Rearrange the equation: 2y’’(exp(3x)) + 3y’(exp(3x)) + 5y(exp(3x)) = x(exp(3x)) 5. Integrate both sides with respect to x: - ∫2y’’(exp(3x))dx + ∫3y’(exp(3x))dx + ∫5y(exp(3x))dx = ∫x(exp(3x))dx

Slide 9

  • Solution for the ODE 2y’’ + 3y’ + 5y = x (contd.): 6. Integrate both sides with respect to x (contd.): - 2∫y’’(exp(3x))dx + 3∫y’(exp(3x))dx + 5∫y(exp(3x))dx = ∫x(exp(3x))dx 7. Simplify and solve the integrals: - 2y’(exp(3x)) + 3y(exp(3x)) + 10y(exp(3x)) = ∫x(exp(3x))dx

Slide 10

  • Solution for the ODE 2y’’ + 3y’ + 5y = x (contd.): 8. Solve for y: - 2y’(exp(3x)) + 13y(exp(3x)) = ∫x(exp(3x))dx 9. Simplify and solve the integral: - 2y’(exp(3x)) + 13y(exp(3x)) = (1/3)x(exp(3x)) - (1/9)exp(3x) + C

Slide 11

  • Solution for the ODE 2y’’ + 3y’ + 5y = x (contd.): 10. Separate variables: - 2y’/y + 13y = (1/3)x - (1/9)exp(3x) + C 11. Integrate both sides: - 2∫y’/y dx + 13∫y dx = ∫(1/3)x - (1/9)exp(3x) dx

Slide 12

  • Solution for the ODE 2y’’ + 3y’ + 5y = x (contd.): 12. Integrate both sides (contd.): - 2ln|y| + 13y = (1/6)x^2 - (1/27)exp(3x) + C1x + C2 13. Solve for y: - 2ln|y| + 13y = (1/6)x^2 - (1/27)exp(3x) + C1x + C2

Slide 13

  • Solution for the ODE 2y’’ + 3y’ + 5y = x (contd.): 14. Solve for y (contd.): - 2ln|y| + 13y = (1/6)x^2 - (1/27)exp(3x) + C1x + C2 15. This equation cannot be solved explicitly for y 16. The general solution is given by: - y = exp(-13/2x) * (C3sin(kx) + C4cos(kx))

Slide 14

  • Further Analysis of the Solution:
    1. The solution contains two arbitrary constants, C3 and C4
    2. The values of C3 and C4 can be determined using initial conditions or boundary conditions
    3. The solution is in terms of sin(kx) and cos(kx), where k is a constant
    4. The value of k determines the nature of the solution

Slide 15

  • Types of Solutions:
    1. Oscillatory Solution:
      • When k is a real value, the solution is oscillatory
      • Example: y = exp(-13/2x) * (C3sin(kx) + C4cos(kx))
    2. Exponential Growth or Decay:
      • When k is a complex value, the solution exhibits exponential growth or decay
      • Example: y = exp(-13/2x) * (C3sin(kx) + C4cos(kx))

Slide 16

  • Example: Solve the ODE 3xy’ - y = x
    1. Rearrange the equation: y’ - (1/3x)y = (1/3)x^2
    2. Identify p(x): p(x) = -1/(3x)
    3. Find the integrating factor: IF = exp(∫-1/(3x)dx) = exp(-1/3 ln|x|) = |x|^(-1/3)

Slide 17

  • Example: Solve the ODE 3xy’ - y = x (contd.) 4. Multiply the equation by the integrating factor: |x|^(-1/3) y’ - (1/3) |x|^(-1/3) y = (1/3) x^(5/3) 5. Simplify and rearrange the equation: (|x|^(-1/3) y)’ = (1/3) x^(5/3) 6. Integrate both sides with respect to x: ∫(|x|^(-1/3) y)’ dx = ∫(1/3) x^(5/3) dx

Slide 18

  • Example: Solve the ODE 3xy’ - y = x (contd.) 7. Integrate both sides with respect to x (contd.): |x|^(-1/3) y = (1/3) (3/8) x^(8/3) + C₁ 8. Solve for y: y = (8/9) x + C₁ |x|^(1/3)
  • Note: The absolute value of x is used to address both positive and negative values of x

Slide 19

  • Summary:
    1. Ordinary Differential Equations (ODEs) can be linear or nonlinear
    2. Linear ODEs can be homogeneous or non-homogeneous
    3. Linear ODEs with constant coefficients can be solved using the method of Integrating Factors
    4. The general solution of a linear ODE contains arbitrary constants
    5. The values of these constants can be determined using initial conditions or boundary conditions
    6. The nature of the solution depends on the value of the constant ‘k’ in the general solution

Slide 20

  • Review questions:
    1. What is the difference between a homogeneous and non-homogeneous linear ODE?
    2. How do you solve a linear ODE with constant coefficients using the method of Integrating Factors?
    3. How do you determine the values of the arbitrary constants in the general solution of a linear ODE?
    4. What is the nature of the solution for a linear ODE when ‘k’ is a real value? What about when it is a complex value?
    5. Can you solve the ODE 3xy’ - y = x and find the general solution?

Slide 21

  • Definition and Role of Integration in Calculus
  • Indefinite and Definite Integrals
  • Relationship between Derivatives and Integrals
  • Fundamental Theorem of Calculus

Slide 22

  • Integration Techniques:
    • Substitution Method
    • Integration by Parts
    • Partial Fractions
    • Trigonometric Substitutions
    • Improper Integrals

Slide 23

  • Examples of Integration Techniques:
    • Substitution: ∫(2x + 3)^4 dx
    • Integration by Parts: ∫x sin(x) dx
    • Partial Fractions: ∫(x + 1)/(x^2 + x) dx
    • Trigonometric Substitutions: ∫(1 - x^2)^(3/2) dx
    • Improper Integrals: ∫(0 to ∞) e^(-x) dx

Slide 24

  • Applications of Integration:
    • Area under a Curve
    • Finding the Volume of Solids of Revolution
    • Calculation of Arc Length
    • Surface Area
    • Work and Energy

Slide 25

  • Example of Application: Area under a Curve
    • Find the area bounded by the curve y = x^2, x-axis, and the lines x = 1 and x = 2

Slide 26

  • Example of Application: Volume of Solids of Revolution
    • Find the volume of the solid generated by rotating the region bounded by the curve y = x^2 and the line y = 0 about the x-axis

Slide 27

  • Example of Application: Calculation of Arc Length
    • Find the length of the curve y = x^2 from x = 0 to x = 2

Slide 28

  • Example of Application: Surface Area
    • Find the surface area of the solid generated by rotating the curve y = x^2 from x = 0 to x = 2 about the x-axis

Slide 29

  • Example of Application: Work and Energy
    • Find the work done by the force F(x) = 3x on an object moving from x = 1 to x = 4

Slide 30

  • Summary:
    • Integration plays a fundamental role in Calculus
    • There are various techniques for integration, including substitution, integration by parts, partial fractions, trigonometric substitutions, and improper integrals
    • Integration has applications in calculating areas, volumes, arc length, surface area, and work and energy
    • Understanding and mastering integration is crucial for solving a wide range of mathematical problems