First-order differential equations are the building blocks of calculus-based modeling. They describe how things change over time, like population growth or chemical reactions. These equations involve the first derivative of an unknown function and can be written as dy/dx = f(x, y).

Understanding how to classify and solve these equations is crucial. We'll look at separable, linear, exact, and homogeneous types, each with its own solving method. This knowledge lets us tackle real-world problems and predict future outcomes in various fields.

Classifying Differential Equations

First-Order Differential Equations

• First-order differential equations involve the first derivative of an unknown function
• Can be written in the general form $\frac{dy}{dx} = f(x, y)$
• The order of a differential equation is determined by the highest derivative present in the equation
• Examples of first-order differential equations:
  • $\frac{dy}{dx} = x^2 + y^2$
  • $\frac{dy}{dx} + 2xy = e^x$

Types of First-Order Differential Equations

• First-order differential equations can be classified based on their structure and the relationship between variables and derivatives
• Common types include:
  • Separable: Can be written in the form $\frac{dy}{dx} = f(x)g(y)$, where the right-hand side can be factored into a function of $x$ and a function of $y$
  • Linear: Can be written in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$ only
  • Exact: The equation $M(x, y)dx + N(x, y)dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
  • Homogeneous: Can be written in the form $\frac{dy}{dx} = f(\frac{y}{x})$, where the right-hand side is a function of the ratio $\frac{y}{x}$
• The classification of a first-order differential equation determines the appropriate solution method to be used, such as separation of variables, integrating factors, or substitution

Solving Separable and Linear Equations

Solving Separable Differential Equations

• To solve a separable differential equation, separate the variables by dividing both sides by $g(y)$ and multiplying by $dx$
• Integrate both sides with respect to their corresponding variables
• Example: Solve $\frac{dy}{dx} = xy$
  • Separate variables: $\frac{1}{y}dy = xdx$
  • Integrate: $\ln|y| = \frac{1}{2}x^2 + C$
  • Solve for $y$: $y = \pm e^{\frac{1}{2}x^2 + C}$

Solving Linear First-Order Differential Equations

• The integrating factor method is used to solve linear first-order differential equations
• Multiply both sides of the equation by an integrating factor, which is $e^{\int P(x)dx}$
• After applying the integrating factor, the left-hand side becomes the derivative of a product, allowing for integration to find the general solution
• Example: Solve $\frac{dy}{dx} + 2y = e^x$
  • Integrating factor: $\mu(x) = e^{\int 2dx} = e^{2x}$
  • Multiply both sides by $\mu(x)$: $e^{2x}\frac{dy}{dx} + 2e^{2x}y = e^{3x}$
  • Simplify: $\frac{d}{dx}(e^{2x}y) = e^{3x}$
  • Integrate: $e^{2x}y = \frac{1}{3}e^{3x} + C$
  • Solve for $y$: $y = \frac{1}{3}e^x + Ce^{-2x}$

Modeling with Differential Equations

Formulating Differential Equations for Real-World Problems

• First-order differential equations can model various real-world phenomena, such as population growth, radioactive decay, cooling/heating, and mixing problems
• To model a real-world problem using a first-order differential equation:
  • Identify the relevant variables, their relationships, and the rates of change involved
  • Express the rate of change of the dependent variable in terms of the independent variable and any other given information
• Example: Formulate a differential equation for population growth with a constant growth rate
  • Let $P(t)$ be the population at time $t$ and $k$ be the constant growth rate
  • The rate of change of population is proportional to the current population: $\frac{dP}{dt} = kP$

Solving and Interpreting Differential Equation Models

• Solve the resulting differential equation using the appropriate method based on its classification
• Apply initial conditions to find the particular solution
• Interpret the solution in the context of the real-world problem and use it to make predictions or draw conclusions
• Example: Solve the population growth differential equation $\frac{dP}{dt} = kP$ with initial condition $P(0) = P_0$
  • Separate variables and integrate: $\int \frac{1}{P}dP = \int kdt$
  • Solve for $P(t)$: $P(t) = P_0e^{kt}$
  • Interpret the solution: The population grows exponentially with a constant growth rate $k$, starting from an initial population $P_0$

Existence and Uniqueness of Solutions

Initial Value Problems (IVPs)

• An initial value problem (IVP) consists of a first-order differential equation and an initial condition specifying the value of the unknown function at a particular point
• Example: $\frac{dy}{dx} = x + y$, $y(0) = 1$
• The existence and uniqueness theorem for first-order IVPs states that if $f(x, y)$ is continuous and satisfies the Lipschitz condition in a region containing the initial point, then the IVP has a unique solution in some interval around the initial point

Lipschitz Condition

• The Lipschitz condition requires that $|f(x, y_1) - f(x, y_2)| \leq L|y_1 - y_2|$ for all $(x, y_1)$ and $(x, y_2)$ in the region, where $L$ is a positive constant called the Lipschitz constant
• The Lipschitz condition ensures that the function $f(x, y)$ does not change too rapidly with respect to $y$, which is necessary for the existence and uniqueness of solutions
• Example: Check if $f(x, y) = x + y$ satisfies the Lipschitz condition
  • $|f(x, y_1) - f(x, y_2)| = |x + y_1 - (x + y_2)| = |y_1 - y_2|$
  • The Lipschitz condition is satisfied with $L = 1$

Determining Existence and Uniqueness

• If the existence and uniqueness conditions are not satisfied, an IVP may have no solution, infinitely many solutions, or solutions that are not unique
• The existence and uniqueness of solutions can be determined by verifying the continuity of $f(x, y)$ and checking the Lipschitz condition in the region of interest
• Example: Determine the existence and uniqueness of solutions for the IVP $\frac{dy}{dx} = \sqrt{|y|}$, $y(0) = 0$
  • $f(x, y) = \sqrt{|y|}$ is continuous for all $(x, y)$
  • However, $f(x, y)$ does not satisfy the Lipschitz condition at $(0, 0)$
  • The IVP has infinitely many solutions of the form $y(x) = \begin{cases} 0, & x \leq c \ \frac{1}{4}(x - c)^2, & x > c \end{cases}$ for any constant $c \geq 0$