Difficulty: Intermediate | Prerequisites: Integration techniques (substitution, partial fractions), basic algebra of logarithms and exponentials.
TL;DR: A separable differential equation can be split so that all y terms are on one side and all x terms on the other. You then integrate both sides. An initial condition pins down the constant of integration, giving a particular solution. Slope fields let you visualise solutions without solving.
Differential equations (DEs) are equations involving a function and its derivatives. In Calculus II, the focus is on first-order separable DEs, which are the simplest type to solve analytically. This is your first real encounter with modelling: DEs describe how a quantity changes over time (population growth, radioactive decay, cooling, mixing problems). Sections 9.1 through 9.4 cover the vocabulary, separation of variables, direction (slope) fields, and specific models like the logistic equation. If you are comfortable integrating using substitution and partial fractions, you have the tools you need.
Differential equation (DE)
An equation that relates a function to one or more of its derivatives. In this course, the equations are first-order (involving only y' = dy/dx, not higher derivatives).
Separable equation
A first-order DE that can be written as dy/dx = f(x) · g(y), so you can move all y terms to one side and all x terms to the other. In simple terms, you can "separate" the variables.
General solution
The family of all solutions, containing an arbitrary constant C. It represents every curve that satisfies the DE.
Particular solution
One specific solution obtained by using an initial condition to determine C.
Initial value problem (IVP)
A DE paired with a condition like y(x₀) = y₀. The initial condition picks out exactly one solution curve from the general family.
Slope field (direction field)
A grid of short line segments showing the slope y' at many points (x, y). Each segment is tangent to the solution curve passing through that point. Think of it as a map that shows which way the solution curves flow.
Logistic equation
The DE y' = ky(1 - y/M) or equivalently y' = y(1 - y) in its normalised form. It models population growth that levels off at a carrying capacity. The solution is a sigmoid (S-shaped) curve.
Equilibrium solution
A constant solution y = c where y' = 0 for all x. For y' = y² - y, the equilibria are y = 0 and y = 1.
Solving separable DEs
Write the equation as dy/dx = f(x) · g(y).
Separate: (1/g(y)) dy = f(x) dx.
Integrate both sides.
Solve for y if possible.
Apply the initial condition (if given) to find C.
Example approach (Problem 12): Find f(x) such that f'(x) = f(x)(1 - f(x)) and f(0) = 1/14.
This is the logistic equation. Let y = f(x). Then dy/dx = y(1 - y).
Separate: dy / [y(1 - y)] = dx.
Use partial fractions: 1/[y(1 - y)] = 1/y + 1/(1 - y).
Integrate: ln|y| - ln|1 - y| = x + C, so ln|y/(1 - y)| = x + C.
Exponentiate: y/(1 - y) = Ae^x where A = e^C.
Apply f(0) = 1/14: (1/14)/(1 - 1/14) = (1/14)/(13/14) = 1/13 = A.
So y/(1 - y) = (1/13)e^x, and solving: y = e^x / (13 + e^x) = 1 / (1 + 13e^(-x)).
Example approach (Problem 13): Curve through (0, 6) with slope x/y.
dy/dx = x/y. Separate: y dy = x dx.
Integrate: y²/2 = x²/2 + C.
Apply (0, 6): 36/2 = 0 + C, so C = 18.
The equation is y² = x² + 36, i.e. y = √(x² + 36) (taking the positive root since y(0) = 6 > 0).
Example approach (Problem 15a): General solution of y' = y² - y.
Separate: dy / [y(y - 1)] = dx.
Partial fractions: 1/[y(y - 1)] = 1/(y - 1) - 1/y.
Integrate: ln|y - 1| - ln|y| = x + C, so ln|(y - 1)/y| = x + C.
Exponentiate and solve for y: (y - 1)/y = Ae^x, giving 1 - 1/y = Ae^x, so y = 1/(1 - Ae^x).
Equilibrium solutions: y = 0 and y = 1 (check by substitution).
Example approach (Problem 15b): IVP y' = 4xy + 3y, y(-2) = 1.
Factor: y' = y(4x + 3). Separate: dy/y = (4x + 3) dx.
Integrate: ln|y| = 2x² + 3x + C.
Apply y(-2) = 1: ln(1) = 2(4) + 3(-2) + C = 8 - 6 + C = 2 + C, so C = -2.
Solution: y = e^(2x² + 3x - 2).
Slope fields (Problem 14)
To match a DE to its slope field, evaluate the sign and magnitude of y' at key points:
y' = 1 + x² + y² is always positive (slopes always go up). This matches a field where every segment points upward.
y' = xe^(-x² - y²) changes sign with x (negative for x < 0, positive for x > 0, zero on the y-axis). Slopes flatten out as |x| and |y| grow because of the exponential decay.
y' = 1/(1 + e^(x² + y²)) is always positive but very small (between 0 and 1/2 for most of the plane). Nearly flat everywhere.
y' = sin(xy)cos(xy) = (1/2)sin(2xy) oscillates between positive and negative depending on the product xy. This produces a pattern of alternating slope signs.
Losing solutions when dividing by y. If you divide both sides by g(y), you may lose equilibrium solutions where g(y) = 0. Always check for constant solutions separately.
Forgetting the absolute value inside the logarithm. When you integrate 1/y, you get ln|y|, not ln(y). The absolute value matters when y could be negative.
Dropping the constant of integration too early. Add C immediately after integrating. Students who wait often forget it entirely.
Confusing general and particular solutions. The general solution has C in it. A particular solution has a specific value of C found from the initial condition. The exam may ask for one or the other.
Misreading slope fields. When matching DEs to slope fields, check specific points (the origin, the axes, large values) rather than trying to eyeball the whole picture at once.
⚠️ Problems 12 through 15 on the review are all DEs, covering separation of variables, the logistic equation, slope field matching, and IVPs. This is a substantial portion of the exam.
⚠️ Partial fractions are required for the logistic-type problems (12, 15a). If your partial fractions are rusty, revise them before the exam.
⚠️ Slope field matching (Problem 14) tests conceptual understanding, not computation. Practise evaluating the sign of y' at a few test points to distinguish between DEs.
⚠️ The exam may give you a DE you have not seen before, but it will be separable. The five-step procedure (separate, integrate, solve, apply IC) is the same every time.
True or false: Every first-order DE is separable. False. Only those that can be written as dy/dx = f(x)g(y).
Fill in the blank: After separating variables and integrating, you apply the ______ to find C. Initial condition.
True or false: The logistic equation y' = y(1 - y) has equilibrium solutions at y = 0 and y = 1. True.
Fill in the blank: In a slope field, each short segment shows the ______ of the solution at that point. Slope (i.e. value of y').
True or false: If y' = xe^(-x² - y²), the slope is zero everywhere on the y-axis. True (because x = 0 there).
Q: (Problem 12) Find f(x) such that f'(x) = f(x)(1 - f(x)) and f(0) = 1/14.
A: Separate and use partial fractions: ∫ dy/[y(1-y)] = ∫ dx gives ln|y/(1-y)| = x + C. With y(0) = 1/14: A = 1/13. Solution: f(x) = 1/(1 + 13e^(-x)).
Q: (Problem 13) Find an equation of the curve through (0, 6) whose slope at (x, y) is x/y.
A: dy/dx = x/y. Separate: y dy = x dx. Integrate: y²/2 = x²/2 + C. Apply (0, 6): C = 18. So y² = x² + 36.
Q: (Problem 14) How do you match y' = 1 + x² + y² to a slope field?
A: Since 1 + x² + y² > 0 everywhere, the slope is always positive. Look for the field where every segment points upward. The slopes also grow rapidly as x or y increases.
Q: (Problem 15a) Find the general solution of y' = y² - y.
A: Factor: y' = y(y - 1). Separate: dy/[y(y-1)] = dx. Partial fractions: 1/(y-1) - 1/y. Integrate: ln|(y-1)/y| = x + C. Solve: y = 1/(1 - Ae^x). Equilibria: y = 0, y = 1.
Q: (Problem 15b) Solve y' = 4xy + 3y, y(-2) = 1.
A: Factor: y' = y(4x + 3). Separate: dy/y = (4x+3) dx. Integrate: ln|y| = 2x² + 3x + C. Apply y(-2) = 1: 0 = 8 - 6 + C, so C = -2. Solution: y = e^(2x² + 3x - 2).
Separable DEs model exponential growth and decay (radioactive half-life, bacterial colonies, compound interest). The logistic equation models populations that hit a ceiling, such as fish in a pond or the spread of a disease through a finite population. Newton's law of cooling (the rate of temperature change is proportional to the difference from ambient temperature) is another separable DE that engineers use when designing thermal systems.
Differential equations connect to everything that comes after in engineering mathematics. In later courses (Calc III, linear algebra, ODE courses), you will encounter non-separable, higher-order, and systems of DEs. The logistic equation from this section reappears in ecology, epidemiology, and machine learning (the logistic/sigmoid function).
Slope fields are a gateway to numerical methods (Euler's method), which approximate solutions when no closed-form answer exists.
Tags: differential equations, separable DE, separation of variables, initial value problem, IVP, general solution, particular solution, logistic equation, slope field, direction field, equilibrium solution, partial fractions in DEs, Calculus II, Calc 2, NYU, sections 9.1 9.2 9.3 9.4, first-order ODE