1) Introduction: What Are Differential Equations? ๐
Welcome to the world of differential equations! These are equations that involve derivatives, describing how things change over time or spaceโlike population growth or cooling coffee. A differential equation relates a function to its derivatives. This topic introduces the basics, focusing on separable and first-order ordinary differential equations (ODEs).
Weโll cover:
- Basics of ODEs: Understanding what they are and their importance.
- Separable Equations: A method to solve certain ODEs by separating variables.
- First-Order ODEs: General forms and solutions, including a brief look at other solution methods.
- Real-world Applications: Seeing how ODEs model dynamic systems.
- A look forward: A brief introduction to higher order and partial differential equations.
Quick Recap: Derivatives measure rates of change, and now weโll use them to model dynamic systems.
2) Basics of Differential Equations ๐
A differential equation involves a function and its derivatives. For example, \( \frac{dy}{dx} = 2x \) is a simple ODE. The order is the highest derivative (first-order here), and it can be ordinary (one variable) or partial (multiple variables, coming later).
Definition 16.1: Differential Equation
A differential equation is an equation that includes derivatives of a function. A solution is a function \( y(x) \) that satisfies the equation when substituted.
Example: \( \frac{dy}{dx} = ky \) (exponential growth model).
Solutions can be general (with constants) or particular (with specific values). Letโs explore a basic case.
Example 1: Verifying a Solution
Verify if \( y = e^{2x} \) solves \( \frac{dy}{dx} = 2y \).
- Differentiate: \( \frac{dy}{dx} = 2e^{2x} \).
- Check: \( 2y = 2e^{2x} \), which matches!
Answer: Yes, itโs a solution.
Example 2: General Solution
Find the general solution to \( \frac{dy}{dx} = 3x^2 \).
- Integrate: \( y = \int 3x^2 \, dx = x^3 + C \).
Answer: \( y = x^3 + C \).
3) Separable Differential Equations ๐
Separable equations are first-order ODEs where variables can be isolated on opposite sides, allowing integration. The form is \( \frac{dy}{dx} = g(x)h(y) \).
Definition 16.2: Separable Equation
A separable ODE can be written as \( \frac{dy}{h(y)} = g(x) \, dx \). Integrate both sides to solve.
Steps: Separate variables, integrate each side, and solve for \( y \).
Example 3: Basic Separation
Solve \( \frac{dy}{dx} = xy \).
- Separate: \( \frac{dy}{y} = x \, dx \).
- Integrate: \( \int \frac{1}{y} \, dy = \int x \, dx \).
- Result: \( \ln|y| = \frac{x^2}{2} + C \).
- Solve: \( y = \pm e^{\frac{x^2}{2} + C} = Ae^{\frac{x^2}{2}} \) (where \( A = \pm e^C \)).
Answer: \( y = Ae^{\frac{x^2}{2}} \).
Example 4: With Initial Condition
Solve \( \frac{dy}{dx} = 2xy \), \( y(0) = 1 \).
- Separate: \( \frac{dy}{y} = 2x \, dx \).
- Integrate: \( \ln|y| = x^2 + C \).
- Exponential: \( y = e^{x^2 + C} = Ae^{x^2} \).
- Apply \( y(0) = 1 \): \( 1 = Ae^0 \), so \( A = 1 \).
Answer: \( y = e^{x^2} \).
4) First-Order ODEs and General Forms ๐
First-order ODEs involve only the first derivative. Separable equations are a subset, but not all are separable. The general form is \( \frac{dy}{dx} = f(x, y) \).
Definition 16.3: First-Order ODE
A first-order ODE is \( \frac{dy}{dx} = f(x, y) \). A solution satisfies the equation for all \( x \) in the domain.
Non-separable ODEs may require other methods (e.g., integrating factors), but weโll focus on separable ones here.
Example 5: Non-Linear Separable
Solve \( \frac{dy}{dx} = \frac{y^2}{x} \).
This is a first-order, non-linear separable differential equation. We'll solve it step-by-step:
-
Separate the variables:
We want to get all terms involving \( y \) on one side and all terms involving \( x \) on the other. Divide both sides by \( y^2 \) and multiply both sides by \( dx \):
\( \frac{1}{y^2} dy = \frac{1}{x} dx \)
-
Integrate both sides:
Integrate both sides of the equation with respect to their respective variables:
\( \int \frac{1}{y^2} dy = \int \frac{1}{x} dx \)
This is equivalent to:
\( \int y^{-2} dy = \int x^{-1} dx \)
Applying the power rule for integration on the left and the natural logarithm rule on the right, we get:
\( -y^{-1} = \ln|x| + C \)
Which simplifies to:
\( -\frac{1}{y} = \ln|x| + C \)
-
Solve for \( y \):
Multiply both sides by -1:
\( \frac{1}{y} = -\ln|x| - C \)
Take the reciprocal of both sides:
\( y = \frac{1}{-\ln|x| - C} \)
We can also rewrite this as:
\( y = -\frac{1}{\ln|x| + C} \)
Answer: \( y = -\frac{1}{\ln|x| + C} \).
Domain and Considerations:
- The natural logarithm \( \ln|x| \) is defined only for \( x \neq 0 \).
- The denominator \( \ln|x| + C \) cannot be zero, which imposes further restrictions on the domain.
- The solution \( y \) cannot be zero.
Example 6: Real-World Application - Logistic Growth
A population grows according to the logistic model \( \frac{dy}{dt} = ry(1 - \frac{y}{K}) \), where \( r \) is the growth rate and \( K \) is the carrying capacity. Solve this with \( y(0) = y_0 \).
This model describes how a population grows exponentially at first, but then slows down as it approaches the carrying capacity of the environment.
- Separate: \( \frac{dy}{y(1 - \frac{y}{K})} = r \, dt \).
- Partial fractions: \( \frac{1}{y(1 - \frac{y}{K})} = \frac{1}{y} + \frac{1}{K-y} \).
- Integrate: \( \int \left( \frac{1}{y} + \frac{1}{K-y} \right) dy = \int r \, dt \).
- Result: \( \ln|y| - \ln|K-y| = rt + C \).
- Simplify: \( \ln \left| \frac{y}{K-y} \right| = rt + C \).
- Exponential: \( \frac{y}{K-y} = Ae^{rt} \) (where \( A = e^C \)).
- Solve: \( y = \frac{KAe^{rt}}{1 + Ae^{rt}} \).
- Apply \( y(0) = y_0 \): \( y_0 = \frac{KA}{1 + A} \), so \( A = \frac{y_0}{K-y_0} \).
Answer: \( y(t) = \frac{Ky_0e^{rt}}{K-y_0 + y_0e^{rt}} \). This solution shows how the population approaches \( K \) as \( t \) increases.
Brief Look at Other Methods: For non-separable first-order ODEs, techniques like integrating factors are used. For example, linear first-order ODEs of the form \( \frac{dy}{dx} + P(x)y = Q(x) \) can be solved using an integrating factor \( e^{\int P(x) \, dx} \). These methods will be covered in more advanced topics.
5) Practice Questions ๐ฏ
Fundamental Practice Questions ๐ฑ
Instructions: Solve the following separable differential equations. ๐
\( \frac{dy}{dx} = 2xy \)
\( \frac{dy}{dx} = x^2 y \)
\( \frac{dy}{dx} = 3y \)
\( \frac{dy}{dx} = \frac{x}{y} \)
\( \frac{dy}{dx} = 4x^3 y^2 \)
\( \frac{dy}{dx} = e^x y \)
\( \frac{dy}{dx} = \frac{1}{x} y \)
\( \frac{dy}{dx} = \sin(x) y \)
\( \frac{dy}{dx} = \frac{y}{1+x^2} \)
\( \frac{dy}{dx} = \cos(x) (1 + y) \)
\( \frac{dy}{dx} = x e^{-y} \)
Challenging Practice Questions ๐
Instructions: Solve these advanced separable ODEs with initial conditions or interpretations. ๐ง
Solve \( \frac{dy}{dx} = \frac{y}{x} \), \( y(1) = 2 \), and verify.
Find the solution to \( \frac{dy}{dx} = ky(1 - \frac{y}{K}) \), \( y(0) = y_0 \), and describe its behavior for different \( y_0 \) values.
Determine \( \frac{dy}{dx} = x y^2 \), \( y(0) = 1 \), and check for singularities.
Evaluate \( \frac{dy}{dx} = \frac{\cos(x)}{y} \), \( y(0) = 1 \), and interpret physically (e.g., related to some physical oscillation).
Find the general solution to \( \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \) and discuss its domain. Consider the solution when \( y(0) = 1 \).
6) Summary & Cheat Sheet ๐
6.1) Differential Equations
An equation involving derivatives; a solution is a function satisfying it.
6.2) Separable ODEs
Form \( \frac{dy}{h(y)} = g(x) \, dx \), integrate both sides.
6.3) First-Order ODEs
General form \( \frac{dy}{dx} = f(x, y) \); separable is a key type. Other methods (integrating factors) exist for non-separable cases.
6.4) Real-World Applications
ODEs model various phenomena like population growth (logistic model) and physical processes.
Youโve begun your differential equations journey! Next, weโll apply them to more real-world problems and explore other types of ODEs. ๐