The Main Idea 

Think of linear differential equations like upgrading from a basic physics problem to the real world—instead of just gravity pulling a ball down, now we add air resistance, friction, and other forces that make the math more interesting but the results more accurate.

What Makes an Equation “Linear”? The unknown function [latex]y[/latex] and its derivative [latex]y'[/latex] appear only to the first power and never get multiplied together. The general form is: [latex]a(x)y' + b(x)y = c(x)[/latex]

Why “Linear” Matters:

• Linear equations have systematic solution methods
• Nonlinear equations often require special techniques or numerical approximation
• Linear systems are more predictable and well-behaved

The Main Idea 

Standard form is like having all your tools organized before starting a project. Once your equation looks like [latex]y' + p(x)y = q(x)[/latex], you can apply systematic solution methods (like integrating factors) that work every time.

What Are [latex]p(x)[/latex] and [latex]q(x)[/latex]?

• [latex]p(x)[/latex] is the coefficient of [latex]y[/latex]
• [latex]q(x)[/latex] is everything on the right-hand side
• Both depend only on [latex]x[/latex], not on [latex]y[/latex]

Problem-Solving Strategy:

1. Rearrange: Move all terms with [latex]y[/latex] or [latex]y'[/latex] to the left side
2. Factor: Make sure [latex]y'[/latex] and [latex]y[/latex] terms are separated
3. Normalize: Divide everything by the coefficient of [latex]y'[/latex]

Quick Examples:

• [latex]y' = 3x - 4y[/latex] → [latex]y' + 4y = 3x[/latex] (just rearrange)
• [latex]3y' - 6y = 9x[/latex] → [latex]y' - 2y = 3x[/latex] (divide by [latex]3[/latex])
• [latex]xy' + 2y = x^2[/latex] → [latex]y' + \frac{2}{x}y = x[/latex] (divide by [latex]x[/latex])

The Main Idea 

Think of an integrating factor as the special function that makes the left side of your equation collapse into something you can actually work with. The integrating factor [latex]\mu(x) = e^{\int p(x)dx}[/latex] is designed to turn the messy left side [latex]y' + p(x)y[/latex] into the clean derivative [latex]\frac{d}{dx}[\mu(x)y][/latex]. This transforms your differential equation into something you can integrate directly.

Problem-Solving Strategy:

• Step 1: Get your equation into standard form [latex]y' + p(x)y = q(x)[/latex]
• Step 2: Calculate [latex]\mu(x) = e^{\int p(x)dx}[/latex] (no constant needed here!)
• Step 3: Multiply the entire equation by [latex]\mu(x)[/latex]
• Step 4: The left side becomes [latex]\frac{d}{dx}[\mu(x)y][/latex], so integrate both sides
• Step 5: Solve for [latex]y[/latex] and apply initial conditions if given

After multiplying by [latex]\mu(x)[/latex], the left side always becomes [latex]\frac{d}{dx}[\mu(x)y] = \mu(x)q(x)[/latex]. This isn’t magic—it’s the product rule working in reverse.

The Main Idea 

Population growth models evolve from simple exponential growth to more realistic patterns that account for environmental limits. The derivative [latex]\frac{dP}{dt}[/latex] represents how fast a population is changing at any given moment.

Exponential Growth: The equation [latex]\frac{dP}{dt} = rP[/latex] creates unlimited J-shaped growth where bigger populations grow faster. This works well for small populations with abundant resources, but it predicts unrealistic infinite growth over time.

Logistic Growth Model: The equation [latex]\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)[/latex] creates realistic S-shaped growth curves by incorporating carrying capacity.

How the Logistic Model Behaves:

• When [latex]P[/latex] is small: [latex]\frac{P}{K} \approx 0[/latex], so growth looks nearly exponential
• As [latex]P[/latex] approaches [latex]K[/latex]: The term [latex]\left(1 - \frac{P}{K}\right)[/latex] shrinks, slowing growth
• When [latex]P = K[/latex]: Growth rate becomes zero (equilibrium)
• When [latex]P > K[/latex]: Growth rate turns negative, population decreases toward [latex]K[/latex]

Real populations face constraints like limited food, space, and resources. This leads to the concept of carrying capacity [latex]K[/latex]—the maximum population an environment can sustain long-term. Carrying capacity acts like a population magnet—populations below [latex]K[/latex] grow toward it, populations above [latex]K[/latex] decline toward it. This creates the characteristic S-curve that starts exponential, slows down, and levels off at the carrying capacity.

The Main Idea 

The logistic differential equation [latex]\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)[/latex] can be solved exactly using separation of variables, giving us the complete formula for realistic population growth.

The solution process involves separating variables to get [latex]\frac{dP}{P(K-P)} = \frac{r}{K}dt[/latex], then using partial fraction decomposition on the left side: [latex]\frac{K}{P(K-P)} = \frac{1}{P} + \frac{1}{K-P}[/latex]. After integration and algebraic manipulation, we arrive at the general solution.

The Complete Solution: [latex]P(t) = \frac{P_0 K e^{rt}}{(K-P_0) + P_0 e^{rt}}[/latex]

Key Behaviors of This Solution:

• When [latex]t \to -\infty[/latex]: [latex]P(t) \to 0[/latex] (population approaches zero in distant past)
• When [latex]t \to \infty[/latex]: [latex]P(t) \to K[/latex] (population approaches carrying capacity)
• The curve is S-shaped, starting slow, accelerating, then leveling off

The Inflection Point: The population growth changes from accelerating to decelerating at [latex]t = \frac{1}{r}\ln\frac{K-P_0}{P_0}[/latex], which occurs when the population reaches exactly [latex]P = \frac{K}{2}[/latex] (half the carrying capacity). This is where the growth rate is fastest.

Equilibrium Solutions: [latex]P = 0[/latex] (extinction) and [latex]P = K[/latex] (carrying capacity) are constant solutions. The first is unstable (small populations grow away from zero), while the second is stable (populations approach the carrying capacity).