Slide 1

• Topic: Integral Calculus - Homogeneous Ordinary Differential Equation

Slide 2

• Homogeneous Ordinary Differential Equation (ODE) is of the form:
  • dy/dx = f(x, y)

Slide 3

• The solution to a homogeneous ODE is obtained by using a substitution.
• Let y = vx, where v is a function of x.

Slide 4

• Differentiating y = vx, we get:
  • dy/dx = v + x * (dv/dx)

Slide 5

• Substituting the expressions for dy/dx and y in the original ODE, we get:
  • v + x * (dv/dx) = f(x, vx)

Slide 6

• Dividing the equation by x, we get:
  • (1/x)*(dv/dx) = (f(x, vx) - v)/x

Slide 7

• Simplifying the equation, we get:
  • (1/v)*(dv/dx) = (f(x, vx) - v)/x

Slide 8

• Now, let (dv/dx)/v = g(x).
• We can rewrite the equation as:
  • (dv/dx) = v * g(x)

Slide 9

• Integrating both sides of the equation, we get:
  • ∫(dv/v) = ∫g(x) dx

Slide 10

• Solving the integral, we get:
  • ln|v| = ∫g(x) dx + C
  • Where C is the constant of integration.

Slide 11

• To find the value of v, we can exponentiate both sides of the equation:
  • |v| = e^(∫g(x) dx + C)
  • We can drop the absolute value sign since e^(∫g(x) dx + C) is always positive.

Slide 12

• Therefore, v = e^(∫g(x) dx + C)
• Substituting y = vx, we get:
  • y = x * e^(∫g(x) dx + C)

Slide 13

• Simplifying the expression, we get:
  • y = x * e^(∫g(x) dx) * e^C
  • Since e^C is a constant, we can write it as k.

Slide 14

• Therefore, the general solution to the homogeneous ODE is:
  • y = k * x * e^(∫g(x) dx)

Slide 15

• Now let’s look at an example to understand the concept better.
• Consider the ODE: xy’ - y = x^2

Slide 16

• Dividing both sides of the equation by x, we get:
  • y’ - (y/x) = x

Slide 17

• Comparing the equation with dy/dx = f(x, y), we get:
  • f(x, y) = (y/x) - x

Slide 18

• Now we will substitute y = vx, where v is a function of x.
• Differentiating y = vx, we get:
  • dy/dx = v + x * (dv/dx)

Slide 19

• Substituting the expressions for dy/dx and y in the original ODE, we get:
  • v + x * (dv/dx) - v = x
  • Simplifying the equation, we get:
  • x * (dv/dx) = x
  • Dividing both sides by x, we get:
  • (dv/dx) = 1

Slide 20

• Integrating both sides of the equation, we get:
  • ∫dv = ∫1 dx
  • v = x + C
  • Substituting v = x + C in y = vx, we get the general solution to the ODE:
  • y = (x + C) * x

Slide 21

• Let’s take a look at an example to solve a homogeneous ODE using the substitution method.
• Consider the ODE: x(dy/dx) + y = 2x^2 * ln(x)

Slide 22

• Dividing both sides of the equation by x, we get:
  • dy/dx + (y/x) = 2x * ln(x)

Slide 23

• Comparing the equation with dy/dx = f(x, y), we get:
  • f(x, y) = (y/x) - 2x * ln(x)

Slide 24

• Now we will substitute y = vx, where v is a function of x.
• Differentiating y = vx, we get:
  • dy/dx = v + x * (dv/dx)

Slide 25

• Substituting the expressions for dy/dx and y in the original ODE, we get:
  • v + x * (dv/dx) + (vx/x) - 2x * ln(x) = 0

Slide 26

• Simplifying the equation, we get:
  • x * (dv/dx) + v - 2x * ln(x) = 0
  • Dividing both sides by x, we get:
  • (dv/dx) + (v/x) - 2 * ln(x) = 0

Slide 27

• Comparing the equation with dy/dx = f(x, v), we get:
  • f(x, v) = - (v/x) + 2 * ln(x)

Slide 28

• Now we can solve the integral:
  • ∫((-v)/x + 2 * ln(x)) dx = ∫0 dv

Slide 29

• Integrating the equation, we get:
  • -v * ln(x) + 2x * ln(x) - 2x = C
  • Where C is the constant of integration.

Slide 30

• Substituting v = y/x, we get:
  • -y * ln(x) + 2x * ln(x) - 2x = C*x
  • Simplifying the equation, we get the general solution to the ODE:
  • y = C*x^2 * e^(2lnx - xln(x))