Chapter Study Guide: First-order Linear Equations

Learning Objectives

After completing this study guide, you should be able to:

• Identify and write first-order linear differential equations in standard form.
• Calculate the integrating factor for any linear first-order differential equation.
• Apply the integrating factor method to solve general first-order linear differential equations.
• Solve initial-value problems (IVPs) involving first-order linear equations.
• Model applied real-world problems (such as air resistance and electrical circuits) using first-order linear equations.

---

Key Terms & Glossary

• First-order Linear Differential Equation: An equation involving an unknown function $y and its first derivative y'$, where both $y$ and $y' appear linearly (no powers, no products of y$ and $y'$).
• Standard Form: The specific arrangement of a linear differential equation written as $y' + p(x)y = q(x), where the coefficient of y'$ is exactly 1.
• Integrating Factor: A specially determined mathematical multiplier, denoted $I(x)$ or $\mu(x)$, that transforms the left-hand side of a standard-form differential equation into the exact derivative of a product.
• General Solution: A solution that encompasses all possible particular solutions, typically containing an arbitrary constant $C$.
• Initial-Value Problem (IVP): A differential equation paired with a specific starting condition (e.g., $y(x_0) = y_0), allowing you to solve for the exact value of the constant C$.

---

The "Big Idea"

[!TIP] The Reverse Product Rule The "Big Idea" behind solving first-order linear differential equations is a clever trick to exploit the Product Rule for derivatives: $\frac{d}{dx}[u \cdot v] = u'v + uv'$.

By writing the equation in standard form $\frac{dy}{dx} + p(x)y = q(x)$ and multiplying every term by an integrating factor, the left side magically collapses into the derivative of a single product. Once collapsed, you simply integrate both sides to solve for $y$ without the derivatives getting in the way!

---

Formula / Concept Box

---

Hierarchical Outline

1. Basics of First-order Linear Equations
   • Recognizing linearity (no $y^2$, $\sin(y)$, or $y \cdot y'$)
   • The Standard Form: $\frac{dy}{dx} + p(x)y = q(x)$
2. The Integrating Factor Method
   • Deriving the Integrating Factor: $\mu(x) = e^{\int p(x) dx}$
   • Applying the Product Rule in reverse
   • Integrating both sides to find the General Solution
3. Initial-Value Problems (IVPs)
   • Substituting $x_0$ and $y_0$ into the General Solution
   • Solving for the integration constant $C$
4. Applications
   • Modeling population growth/decay
   • Electrical circuits (RL circuits)
   • Newton's Law of Cooling & Air Resistance models

---

Comparison Tables

[!NOTE] Distinguishing between equation types is critical on exams. Use this table to decide your approach.

---

Visual Anchors

Method Flowchart

Solution Curves (Family of Solutions vs. IVP)

[!TIP] In the graph above, all curves represent the General Solution $y = 2 + Ce^{-x}. The single blue dot at (0, 0.5) anchors an **Initial-Value Problem**, isolating exactly one valid curve (the blue line, where C = -1.5$).

---

Definition-Example Pairs

▶Standard Form

Definition: A differential equation formatted such that $\frac{dy}{dx} has a coefficient of exactly 1, followed by a term involving y, set equal to a function of x$.

Example: The equation $x^2 y' + 2xy = e^x is *not* in standard form. Divide by x^2 to get the standard form: \frac{dy}{dx} + \frac{2}{x}y = \frac{e^x}{x^2}$. Here, $p(x) = \frac{2}{x}$ and $q(x) = \frac{e^x}{x^2}$.

▶Integrating Factor

Definition: The term used to multiply both sides of a standard form differential equation, making the left side a perfect derivative. Calculated as $\mu(x) = e^{\int p(x) dx}$.

Example: For the equation $y' + 3y = 6x$, the function $p(x) = 3. The integrating factor is \mu(x) = e^{\int 3 dx} = e^{3x}$.

---

Worked Examples

Example 1: Finding the General Solution

Problem: Find the general solution to the differential equation $x \frac{dy}{dx} + 3y = 4x^2$ assuming $x > 0$.

Step 1: Put into standard form. Divide both sides by $x to make the coefficient of y'$ equal to 1: $\frac{dy}{dx} + \frac{3}{x}y = 4x$ Identify terms: $p(x) = \frac{3}{x}$ and $q(x) = 4x$.

Step 2: Find the integrating factor $\mu(x)$. $\mu(x) = e^{\int p(x) dx} = e^{\int \frac{3}{x} dx}$ $\mu(x) = e^{3 \ln x} = e^{\ln(x^3)} = x^3$

Step 3: Multiply the standard form by $\mu(x)$. $x^3 \left( \frac{dy}{dx} + \frac{3}{x}y \right) = x^3(4x)$ $x^3 \frac{dy}{dx} + 3x^2 y = 4x^4$

Step 4: Collapse the left side. Recognize that the LHS is the exact derivative of $(\mu(x)y)$: $\frac{d}{dx}\left( x^3 y \right) = 4x^4$

Step 5: Integrate both sides. $x^3 y = \int 4x^4 dx$ $x^3 y = \frac{4}{5}x^5 + C$

Step 6: Solve for $y$. $y(x) = \frac{4}{5}x^2 + \frac{C}{x^3}$

Example 2: Solving an Initial-Value Problem (IVP)

Problem: Solve the initial-value problem: $y' + 2y = e^x, \quad y(0) = 2$

Step 1 & 2: Standard form and Integrating factor. The equation is already in standard form where $p(x) = 2$. $\mu(x) = e^{\int 2 dx} = e^{2x}$

Step 3 & 4: Multiply and collapse. $e^{2x}y' + 2e^{2x}y = e^{2x}e^x$ $\frac{d}{dx}(e^{2x}y) = e^{3x}$

Step 5 & 6: Integrate and solve for General Solution. $e^{2x}y = \int e^{3x} dx = \frac{1}{3}e^{3x} + C$ $y(x) = \frac{1}{3}e^x + Ce^{-2x}$

Step 7: Apply the initial condition to find $C$. Substitute $x = 0, y = 2$: $2 = \frac{1}{3}e^0 + Ce^0$ $2 = \frac{1}{3} + C \implies C = \frac{5}{3}$

Final Particular Solution: $y(x) = \frac{1}{3}e^x + \frac{5}{3}e^{-2x}$

---

Checkpoint Questions

Test your active recall. Cover the answers to see if you mastered the concepts!

1. Question: Why is $y^2 y' + 2x = 0$ NOT a linear differential equation?
   • Answer: It contains a $y^2$ term multiplied by $y'. A linear differential equation must be degree 1 with respect to y$ and $y'$ and cannot have them multiplied together.
2. Question: When converting to standard form, what must you do if your equation looks like $\cos(x)y' + y = x$?
   • Answer: You must divide the entire equation by $\cos(x) to ensure the coefficient of y' is exactly 1. The new p(x)$ will be $\frac{1}{\cos(x)} = \sec(x)$.
3. Question: What happens to the $+C when computing the integral for the integrating factor \mu(x)$?
   • Answer: We ignore it (or set $C=0). Any integrating factor will work to make the left side a perfect derivative, so we choose the simplest one where C=0$.
4. Question: After multiplying by the integrating factor, what should the left-hand side of your equation always turn into?
   • Answer: The exact derivative of the product of your integrating factor and $y$, written as $\frac{d}{dx}[\mu(x)y]$.

---

Muddy Points & Cross-Refs

[!WARNING] Common Pitfall: Don't forget to multiply the right-hand side $q(x) by the integrating factor! Many students successfully collapse the left side but accidentally leave q(x)$ unchanged, leading to an incorrect final integral.

• Cross-Reference: If the integral $\int \mu(x)q(x) dx$ is difficult, you may need to apply Integration by Parts or U-Substitution (review Calculus II Techniques of Integration).
• Cross-Reference: The application of integrating factors is deeply tied to Newton's Law of Cooling and Exponential Decay models. Keep these real-world analogs in mind as you solve.