Day 20: Differential Equations, Slope Fields & Exponential Models

Separable Differential Equations

A differential equation relates a function to its derivatives. When the $y$ terms can be pushed to one side and the $x$ terms to the other, you integrate both sides to find the general solution, then use an initial condition to pin down the constant.

Core Theorem

If

\frac{dy}{dx} = f(x)\,g(y)

, rewrite it as

\frac{dy}{g(y)} = f(x)\,dx

and integrate both sides:

\int \frac{dy}{g(y)} = \int f(x)\,dx

Add a single

+C

on the

x

side, then apply the initial condition to solve for

C

.

Step-by-Step SOP

1
Separate
Get all $y$ 's with $dy$ on one side, all $x$ 's with $dx$ on the other.
2
Integrate
Integrate both sides; put a single $+C$ on the $x$ side.
3
Apply the initial condition
Substitute the given point to solve for $C$ .
4
Solve for y
Isolate $y$ explicitly if the problem asks for $y = \dots$

Practice Exercises

Example 01Medium

Solve

\frac{dy}{dx} = \frac{x}{y}

with

y(0) = 3

.

NEED A HINT?

Multiply both sides by

y

and by

dx

to separate the variables.

SHOW DETAILED EXPLANATION

Separate

y\,dy = x\,dx

.

Integrate both sides

\frac{y^2}{2} = \frac{x^2}{2} + C

.

Apply the initial condition

y(0) = 3

:

\frac{9}{2} = 0 + C \implies C = \frac{9}{2}

.

Solve for y

y^2 = x^2 + 9 \implies y = \sqrt{x^2 + 9}

(positive root, since

y(0) = 3 > 0

).

Example 02Medium

Solve

\frac{dy}{dx} = 2xy

with

y(0) = 4

.

NEED A HINT?

Divide both sides by

y

to separate, then integrate.

SHOW DETAILED EXPLANATION

Separate and integrate

\frac{dy}{y} = 2x\,dx \implies \ln|y| = x^2 + C

.

Exponentiate

y = e^{x^2 + C} = A\,e^{x^2}

.

Apply the initial condition

y(0) = 4 \implies A = 4

, so

y = 4e^{x^2}

.

Slope Fields

A slope field draws a short segment at each grid point $(x, y)$ with slope $\frac{dy}{dx}$ evaluated there. Solution curves follow the flow of these segments, so you can sketch a solution even when the equation cannot be solved by hand.

Core Theorem

Given

\frac{dy}{dx} = F(x, y)

, the solution curve through a point

(x, y)

has slope

F(x, y)

there. Every solution curve is tangent to the field segment at each point it passes through.

Practice Exercises

Example 01Easy

For

\frac{dy}{dx} = x + y

, find the slope of the solution curve at the points

(1, 2)

and

(0, 0)

.

NEED A HINT?

Substitute each point into

x + y

.

SHOW DETAILED EXPLANATION

At the point (1, 2)

slope

= 1 + 2 = 3

.

At the point (0, 0)

slope

= 0 + 0 = 0

, so the segment is horizontal.

Example 02Medium

For

\frac{dy}{dx} = \frac{x}{y}

, where are the field segments horizontal, and where are they vertical?

NEED A HINT?

Horizontal when the slope is

0

; vertical (undefined) when the denominator is

0

.

SHOW DETAILED EXPLANATION

Horizontal segments

\frac{x}{y} = 0

when

x = 0

(the

y

-axis), as long as

y \neq 0

.

Vertical segments

The slope is undefined when

y = 0

(the

x

-axis), as long as

x \neq 0

.

Common Pitfalls

⚠
Reading a slope field as a function graphThe field shows slopes at points in the plane, not the graph of $\frac{dy}{dx}$ . A solution curve threads through the field tangentially — it is not one of the segments.

Exponential Growth and Decay

When a quantity changes at a rate proportional to its current amount — populations, radioactive decay, compound interest, cooling — it follows $y = y_0 e^{kt}$ . This is the single most tested differential equation on the AP exam.

Core Theorem

If

\frac{dy}{dt} = ky

, then

y(t) = y_0\,e^{kt}

where

y_0 = y(0)

is the initial amount.

k > 0

gives growth,

k < 0

gives decay. The doubling time is

\frac{\ln 2}{k}

; the half-life is

\frac{\ln 2}{|k|}

.

Step-by-Step SOP

1
Set up the model
Write $y = y_0 e^{kt}$ and read $y_0$ from the initial condition.
2
Solve for k
Use a second data point (or a half-life / doubling time) and take a natural log.
3
Answer the question
Plug in the requested $t$ for an amount, or solve for $t$ for a time.

Practice Exercises

Example 01Medium

A bacteria culture starts with 500 cells and grows to 800 cells after 2 hours. How many cells are there after 5 hours?

NEED A HINT?

Find

k

from the 2-hour data point, then evaluate at

t = 5

.

SHOW DETAILED EXPLANATION

Write the model

y(t) = 500\,e^{kt}

.

Find k

800 = 500\,e^{2k} \implies e^{2k} = 1.6 \implies k = \frac{\ln 1.6}{2} \approx 0.235

.

Evaluate at t = 5

y(5) = 500\,e^{5(0.235)} = 500\,e^{1.175} \approx 1619

cells.

Example 02Medium

A radioactive isotope has a half-life of 10 days. What fraction remains after 25 days?

NEED A HINT?

The half-life gives

k

:

\tfrac{1}{2} = e^{10k}

.

SHOW DETAILED EXPLANATION

Find k from the half-life

\tfrac{1}{2} = e^{10k} \implies k = \frac{\ln(1/2)}{10} = \frac{-\ln 2}{10} \approx -0.0693

.

Evaluate the fraction at t = 25

\frac{y(25)}{y_0} = e^{25k} = e^{-1.733} \approx 0.177

.

Interpret

About

17.7\%

of the original isotope remains after 25 days.

Common Pitfalls

⚠
Confusing k with a percentage rateIn $y = y_0 e^{kt}$ , $k$ is the continuous rate. A $5\%$ continuous rate is $k = 0.05$ ; a $5\%$ rate compounded per period is $k = \ln(1.05) \approx 0.0488$ .
⚠
Sign of k for decayDecay means $k < 0$ . If a decay problem gives you a positive $k$ , recheck the algebra.

Differential Equations And Exponential Models

Separable Differential Equations

Separate

Integrate

Apply the initial condition

Solve for y

Practice Exercises

Separate

Integrate both sides

Apply the initial condition

Solve for y

Separate and integrate

Exponentiate

Apply the initial condition

Slope Fields

Practice Exercises

At the point (1, 2)

At the point (0, 0)

Horizontal segments

Vertical segments

Exponential Growth and Decay

Set up the model

Solve for k

Answer the question

Practice Exercises

Write the model

Find k

Evaluate at t = 5

Find k from the half-life

Evaluate the fraction at t = 25

Interpret

Gary Chang