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

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Solve

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

with

y(0) = 3

.

How to think about it

Get every

y

(with

dy

) on one side and every

x

(with

dx

) on the other, then integrate both sides. Put a single

+C

on the

x

side and use the initial condition to pin it down.

Key idea: Multiply both sides by $y$ and by $dx$ to separate the variables.

Full solution

1. Separate

y\,dy = x\,dx

.

2. Integrate both sides

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

.

3. Apply the initial condition

y(0) = 3

:

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

.

4. Solve for y

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

(positive root, since

y(0) = 3 > 0

).

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 — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Solve

\frac{dy}{dx} = 2xy

with

y(0) = 4

.

Need a hint?

Divide both sides by

y

to separate, then integrate.

Show solution

1. Separate and integrate

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

.

2. Exponentiate

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

.

3. 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.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

For

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

, find the slope of the solution curve at the points

(1, 2)

and

(0, 0)

.

How to think about it

A slope field is just

\frac{dy}{dx}

evaluated on a grid of points: each little segment's slope is what you get by plugging that

(x, y)

into the equation. Nothing is solved — the differential equation itself is the instruction for drawing it.

Key idea: Substitute each point into $x + y$ .

Full solution

1. At the point (1, 2)

slope

= 1 + 2 = 3

.

2. At the point (0, 0)

slope

= 0 + 0 = 0

, so the segment is horizontal.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

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 solution

1. Horizontal segments

\frac{x}{y} = 0

when

x = 0

(the

y

-axis), as long as

y \neq 0

.

2. 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|}

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked 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?

How to think about it

"Rate proportional to amount" is the cue for

y = y_0 e^{kt}

— write that model down at once. You need two numbers:

y_0

from the start, and

k

from a second data point. Solve for

k

before you answer what the question actually asks.

Key idea: Find $k$ from the 2-hour data point, then evaluate at $t = 5$ .

Full solution

1. Write the model

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

.

2. Find k

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

.

3. Evaluate at t = 5

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

cells.

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 — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

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 solution

1. 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

.

2. Evaluate the fraction at t = 25

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

.

3. 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

Worked Examples

1. Separate

2. Integrate both sides

3. Apply the initial condition

4. Solve for y

Separate

Integrate

Apply the initial condition

Solve for y

Practice — Your Turn

1. Separate and integrate

2. Exponentiate

3. Apply the initial condition

Slope Fields

Worked Examples

1. At the point (1, 2)

2. At the point (0, 0)

Practice — Your Turn

1. Horizontal segments

2. Vertical segments

Exponential Growth and Decay

Worked Examples

1. Write the model

2. Find k

3. Evaluate at t = 5

Set up the model

Solve for k

Answer the question

Practice — Your Turn

1. Find k from the half-life

2. Evaluate the fraction at t = 25

3. Interpret

Gary Chang