---
title: "dy/dt = ky — AP Calculus Definition & Exam Guide"
description: "dy/dt = ky is the differential equation for exponential growth and decay, solved by y = y₀e^(kt). Learn how AP Calc Topic 7.8 tests this model on the exam."
canonical: "https://fiveable.me/ap-calc/key-terms/dy-dt-ky"
type: "key-term"
subject: "AP Calculus AB/BC"
unit: "Unit 7"
---

# dy/dt = ky — AP Calculus Definition & Exam Guide

## Definition

dy/dt = ky is the differential equation modeling exponential growth and decay in AP Calculus, stating that a quantity's rate of change is proportional to its current size, where k is the constant of proportionality. With initial condition y = y₀ at t = 0, its solution is y = y₀e^(kt).

## What It Is

dy/dt = ky is the math translation of one sentence: "the [rate of change](/ap-calc/key-terms/rate-of-change "fv-autolink") of a quantity is [proportional](/ap-calc/unit-7/exponential-models-with-differential-equations/study-guide/fzn9urmywhPSE4KUxVcq "fv-autolink") to the size of the quantity" (FUN-7.F.2). The bigger y gets, the faster it changes. That's why bacteria populations explode and radioactive samples fizzle out slowly at the end. The sign of k tells you the direction. If k > 0, you get exponential growth; if k < 0, you get exponential decay.

The payoff is that this equation has a clean [solution](/ap-calc/unit-7/verifying-solutions-for-differential-equations/study-guide/s2nX7AhIBDxGIWwlL82x "fv-autolink") you should know cold. With initial condition y = y₀ when t = 0, the solution is y = y₀e^(kt) (FUN-7.G.1). You can derive it with separation of variables, but on the exam you'll often save time by recognizing the pattern instantly: see "rate proportional to amount," write dy/dt = ky, jump straight to y = y₀e^(kt).

## Why It Matters

This equation is the entire point of Topic 7.8 (Exponential Models with Differential Equations) in [Unit 7](/ap-calc/unit-7 "fv-autolink"). It supports two learning objectives directly. 7.8.A asks you to interpret a differential equation in context, which usually means translating a word problem ("a medication's concentration decreases at a rate proportional to the current concentration") into dy/dt = ky. 7.8.B asks you to actually solve it, producing the [general solution](/ap-calc/key-terms/general-solution "fv-autolink") y = Ce^(kt) or the particular solution y = y₀e^(kt) once you're given an initial condition. It's also one of the few differential equations the CED names explicitly, so it shows up reliably on both the AB and BC exams.

## Connections

### [Exponential Decay (Unit 7)](/ap-calc/key-terms/exponential-decay)

Decay is just dy/dt = ky with k < 0. Radioactive substances and drug concentrations in the bloodstream both follow y = y₀e^(kt) with a negative k, so the quantity shrinks toward zero but never goes negative.

### Separation of Variables (Unit 7)

[Topic 7.7](/ap-calc/unit-7/finding-particular-solutions-using-initial-conditions-separation-variables/study-guide/v0tgJcQJwgznHGMkq2Uy "fv-autolink")'s technique is how you prove the solution. Separate to get dy/y = k dt, integrate both sides to get ln|y| = kt + C, then exponentiate. The e^(kt) doesn't come from nowhere; it falls out of this process.

### General Solution and Initial Condition (Unit 7)

The general solution y = Ce^(kt) describes a whole family of curves. Plugging in the [initial condition](/ap-calc/key-terms/initial-condition "fv-autolink") y = y₀ at t = 0 pins down C = y₀ and gives you the one particular solution that fits your problem.

### The Logistic Model (Unit 7, BC only)

[Topic 7.9](/ap-calc/unit-7/logistic-models-with-differential-equations/study-guide/VWm383QcmHtCJYsFXl0G "fv-autolink") modifies dy/dt = ky by adding a carrying capacity, giving dy/dt = ky(1 − y/a). Exponential growth assumes unlimited resources; logistic growth is what happens when the real world pushes back.

## On the AP Exam

Multiple-choice questions love the translation step. A typical stem describes a real situation, like a bacteria population growing at a rate proportional to the current population, or a radioactive mass decreasing at a rate proportional to its current mass, and asks which differential equation models it. The answer is dy/dt = ky (or dy/dt = −ky if the problem builds the negative sign into the equation). You'll also be asked to go the other way and identify y = y₀e^(kt) as the solution, or to solve for k or t given two data points, which means taking a natural log. On free-response, this model can appear inside a differential equations question where you set up the equation from a verbal description, solve it with an initial condition, and interpret the answer with units. The fastest mistakes to make are writing dy/dt = kt (proportional to time, wrong) and forgetting the initial condition when finding a particular solution.

## dy/dt = ky vs dy/dt = kt

These look almost identical but model completely different behavior. dy/dt = ky says the rate depends on the quantity itself, which gives exponential solutions y = y₀e^(kt). dy/dt = kt says the rate depends on time, which integrates to a quadratic, y = (k/2)t² + C. Read the words carefully. "Proportional to the quantity" means ky; "proportional to time" means kt. MCQ wrong-answer choices are built on exactly this swap.

## Key Takeaways

- dy/dt = ky is the differential equation that comes from the phrase "the rate of change of a quantity is proportional to the size of the quantity."
- With initial condition y = y₀ at t = 0, the solution is y = y₀e^(kt), and you should be able to write that down without re-deriving it every time.
- The sign of k determines everything: k > 0 means exponential growth, k < 0 means exponential decay.
- You can derive the solution using separation of variables, which is why Topic 7.8 comes right after Topic 7.7.
- Don't confuse "proportional to the quantity" (dy/dt = ky) with "proportional to time" (dy/dt = kt); only the first one gives exponential behavior.
- Solving for k or t in y = y₀e^(kt) almost always requires taking a natural log of both sides.

## FAQs

### What is dy/dt = ky in AP Calculus?

It's the differential equation for exponential growth and decay, covered in Topic 7.8. It says the rate of change of a quantity is proportional to the quantity itself, and its solution with initial condition y₀ at t = 0 is y = y₀e^(kt).

### Do I need to memorize the solution to dy/dt = ky, or derive it every time?

Memorize it. The CED (FUN-7.G.1) states the solution form y = y₀e^(kt) directly, so on multiple choice you should jump straight to it. Know the separation-of-variables derivation too, since FRQs can ask you to show the work.

### Is dy/dt = ky the same as the logistic equation?

No. dy/dt = ky assumes unlimited growth and gives a pure exponential. The logistic equation dy/dt = ky(1 − y/a) caps growth at a carrying capacity a, and it's a separate BC-only topic (7.9).

### What does k mean in dy/dt = ky?

k is the constant of proportionality, the relative growth or decay rate. If k > 0 the quantity grows exponentially; if k < 0 it decays. In problems you often solve for k using a second data point, like a half-life, by taking natural logs.

### How is dy/dt = ky different from dy/dt = kt?

dy/dt = ky depends on the quantity y and produces exponential solutions. dy/dt = kt depends on time t and integrates to a quadratic, not an exponential. Exam questions deliberately offer kt as a trap answer when the stem says "proportional to the current amount."

## Related Study Guides

- [7.8 Exponential Models with Differential Equations](/ap-calc/unit-7/exponential-models-with-differential-equations/study-guide/fzn9urmywhPSE4KUxVcq)

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