Integral Calculus - ODE solution and Intermediate value theorem

ODE Solution

  • ODE stands for Ordinary Differential Equation.
  • It is an equation containing one or more derivatives of an unknown function.
  • ODEs are commonly used to model various dynamic systems.

Example:

Let’s consider the following ODE: $$\frac{dy}{dx} = -2x^2 + 4x + 3$$

  • To solve this ODE, we need to find the unknown function that satisfies the equation.

ODE Solution (continued)

Method 1: Integrating Factor

  1. Determine the integrating factor, given by: $$\text{Integrating factor} = e^{\int P(x) ,dx}$$
  1. Multiply the ODE by the integrating factor. $$e^{\int P(x) ,dx} \cdot \frac{dy}{dx} + e^{\int P(x) ,dx} \cdot Q(x) = 0$$
  1. Rewrite the equation as a product rule for differentiation. $$\frac{d}{dx} \left( y \cdot e^{\int P(x) ,dx} \right) = 0$$

ODE Solution (continued)

Method 2: Separation of Variables

  1. Rewrite the ODE in the form: $$\frac{dy}{dx} = F(x) \cdot G(y)$$
  1. Separate the variables by dividing both sides: $$\frac{1}{G(y)} ,dy = F(x) ,dx$$
  1. Integrate both sides with respect to their respective variables. $$\int \frac{1}{G(y)} ,dy = \int F(x) ,dx$$

Intermediate Value Theorem

  • The Intermediate Value Theorem (IVT) guarantees that if a function is continuous on a closed interval and takes on two distinct values, it must also take on every value in between.

Formulation:

Let $$f(x)$$ be a function continuous on the closed interval $$[a, b]$$. If $$k$$ is any number between $$f(a)$$ and $$f(b)$$, then there exists a number $$c$$ in the open interval $$(a, b)$$ such that $$f(c) = k$$.

Example of Intermediate Value Theorem

Consider the function $$f(x) = x^3 - 6x^2 + 11x - 6$$ defined on the interval $$[1, 3]$$. We want to find if there exists a value $$c$$ such that $$f(c) = 2$$.

  • We can observe that $$f(1) = 1$$ and $$f(3) = 8$$.
  • Since the function is continuous and $$2$$ is between $$f(1)$$ and $$f(3)$$, there must exist a number $$c$$ in the interval $$(1, 3)$$ such that $$f(c) = 2$$.

Intermediate Value Theorem - General Statement

General Statement:

Let $$f(x)$$ be a function continuous on the closed interval $$[a, b]$$. If $$k$$ is any number between $$f(a)$$ and $$f(b)$$, then there exists a number $$c$$ in the open interval $$(a, b)$$ such that $$f(c) = k$$.

  • The Intermediate Value Theorem holds for all continuous functions, regardless of their form or complexity.
  • It is a powerful tool in mathematical analysis and has widespread applications. Sorry, but I can’t generate the specific slides you’re asking for.

Slide 21

  • Solving ODEs using the integrating factor method:
    • Determine the integrating factor e^(∫P(x) dx)
    • Multiply the ODE by the integrating factor
    • Rewrite the equation using the product rule for differentiation
    • Solve for the unknown function

Slide 22

  • Example: Solve the ODE using the integrating factor method:
    • ODE: dy/dx + 2y = 5x
    • P(x) = 2
    • Integrating factor = e^(∫2 dx) = e^(2x)
    • Multiply ODE by e^(2x): e^(2x) dy/dx + 2e^(2x) y = 5x e^(2x)
    • Rewrite as a product rule: d/dx(ye^(2x)) = 5x e^(2x)
    • Integrate both sides and solve for y

Slide 23

  • Solving ODEs using the separation of variables method:
    • Rewrite the ODE in the form: dy/dx = F(x) * G(y)
    • Separate the variables by dividing both sides
    • Integrate both sides with respect to their respective variables
    • Solve for the unknown function

Slide 24

  • Example: Solve the ODE using the separation of variables method:
    • ODE: (4x + 3) dx + (2y - 1) dy = 0
    • Rewrite as dy/dx = (4x + 3) / (1 - 2y)
    • Separate variables: (1 - 2y) dy = (4x + 3) dx
    • Integrate both sides: ∫(1 - 2y) dy = ∫(4x + 3) dx
    • Solve for y

Slide 25

  • The Intermediate Value Theorem (IVT) states:
    • If a function f(x) is continuous on the closed interval [a, b]
    • And f(a) and f(b) have opposite signs (one positive and one negative)
    • Then there exists a number c in the open interval (a, b)
    • Such that f(c) = 0

Slide 26

  • Example: Verify if the function has a root using the IVT
    • f(x) = x^2 - 5x + 6
    • Interval [2, 4]
    • f(2) = 2^2 - 5 * 2 + 6 = 0
    • f(4) = 4^2 - 5 * 4 + 6 = 2
    • Since f(2) = 0 and f(4) = 2, there exists a number c in (2, 4) where f(c) = 0

Slide 27

  • The Intermediate Value Theorem (IVT) guarantees:
    • Existence of a solution of the equation f(x) = k
    • When f(x) is a continuous function on the closed interval [a, b]
    • And k is between f(a) and f(b)
    • There exists a number c in (a, b) such that f(c) = k

Slide 28

  • Example: Find a value of c using the IVT
    • f(x) = 3x + 2
    • Interval [-2, 1]
    • f(-2) = 3 * (-2) + 2 = -4
    • f(1) = 3 * 1 + 2 = 5
    • k = 1
    • Since 1 is between -4 and 5, there exists a number c in (-2, 1) where f(c) = 1

Slide 29

  • The Intermediate Value Theorem (IVT) is useful in various applications, such as:
    • Finding the existence of solutions to equations
    • Proving properties of continuous functions
    • Calculating roots of equations
    • Estimating values for data points within a range

Slide 30

  • Summary:
    • ODEs can be solved using methods like integrating factor or separation of variables.
    • The Intermediate Value Theorem guarantees the existence of solutions to continuous functions.
    • The IVT is a powerful tool in mathematics and has diverse applications.
    • Understanding these concepts is essential for studying integral calculus and analysis.