Introduction

In Lesson 1, we learned that a differential equation describes how a quantity changes over time.

Now we face the next question:

How do we actually solve a differential equation?

One of the simplest and most powerful techniques is called Separation of Variables.

This method allows us to solve many important models, including:

• Population growth
• Population decline
• Radioactive decay
• Continuous compound interest
• Bacterial growth
• Inventory growth

These models appear everywhere in science, finance, statistics, and engineering.

---

What Does It Mean to Solve an ODE?

Suppose we have

$$
\frac{dy}{dt}=2t
$$

To solve this equation means:

Find the function whose derivative equals

$$
2t
$$

The answer is

$$
y=t^2+C
$$

where

$$
C
$$

is a constant.

For simple equations this is easy.

However, most differential equations are more complicated.

---

The Idea Behind Separation of Variables

Consider the equation

$$
\frac{dy}{dt}=3y
$$

Notice something important.

The right side depends only on

$$
y
$$

while the left side contains

$$
\frac{dy}{dt}
$$

The idea is to move all terms involving

$$
y
$$

to one side and all terms involving

$$
t
$$

to the other side.

This is called separating the variables.

---

Example 1: Exponential Growth

Consider

$$
\frac{dy}{dt}=3y
$$

---

Step 1: Separate Variables

Divide both sides by

$$
y
$$

to obtain

$$
\frac{1}{y}\frac{dy}{dt}=3
$$

Writing differentials explicitly:

$$
\frac{1}{y}dy=3dt
$$

The variables are now separated.

---

Step 2: Integrate Both Sides

Integrate both sides:

$$\int \frac{1}{y}dy \int 3dt
$$

The left side becomes

$$
\ln|y|
$$

The right side becomes

$$
3t+C
$$

Therefore

$$\ln|y|=3t+C
$$

---

Step 3: Solve for y

Exponentiate both sides:

$$e^{\ln|y|}=e^{3t+C}
$$

which gives

$$|y|=e^Ce^{3t}
$$

Since

$$
e^C
$$

is simply another constant, we write

$$
A=e^C
$$

Thus

$$
y=Ae^{3t}
$$

This is the general solution.

---

Why Is This Called Exponential Growth?

The solution contains

$$
e^{3t}
$$

which grows increasingly rapidly over time.

For example:

t	y (assuming A=1)
0	1
1	20.09
2	403.43
3	8103.08

The growth accelerates.

The larger the value becomes, the faster it grows.

---

Population Growth Model

Suppose

$$
P(t)
$$

represents a population.

A common assumption is:

$$
\frac{dP}{dt}=rP
$$

where

$$
r
$$

is the growth rate.

Solving exactly as before gives

$$
P(t)=P_0e^{rt}
$$

where

$$
P_0
$$

is the initial population.

---

Example: A City Population

Suppose:

$$
P_0=1000
$$

and

$$
r=0.05
$$

Then

$$
P(t)=1000e^{0.05t}
$$

After 10 years:

$$P(10)

1000e^{0.5}
$$

which is approximately

$$
1648.7
$$

The population has grown by almost 65%.

---

Continuous Compound Interest

Suppose money grows continuously.

Let

$$
M(t)
$$

represent the amount of money.

The growth model is

$$
\frac{dM}{dt}=rM
$$

Notice this is identical to the population model.

The solution is

$$
M(t)=M_0e^{rt}
$$

---

Example

Invest

$$
$10,000
$$

at

$$
5%
$$

continuous interest.

Then

$$
M(t)=10000e^{0.05t}
$$

After 20 years:

$$M(20)

10000e^{1}
$$

approximately

$$
$27,183
$$

---

Exponential Decay

Not everything grows.

Many systems shrink over time.

Examples:

• Radioactive materials
• Drug concentrations
• Battery charge
• Machine reliability

Consider

$$
\frac{dy}{dt}=-2y
$$

The negative sign changes everything.

---

Solving

Separate variables:

$$
\frac{1}{y}dy=-2dt
$$

Integrate:

$$\int \frac{1}{y}dy=\int -2dt
$$

giving

$$
\ln|y|=-2t+C
$$

Exponentiating:

$$
y=Ae^{-2t}
$$

---

Why Is This Decay?

Because

$$
e^{-2t}
$$

gets smaller as time increases.

For example:

t	y (assuming A=1)
0	1
1	0.135
2	0.018
3	0.0025

The quantity rapidly approaches zero.

---

Radioactive Decay

A radioactive substance often follows

$$
\frac{dN}{dt}=-\lambda N
$$

where

$$
N
$$

is the amount remaining and

$$
\lambda
$$

is the decay constant.

The solution is

$$
N(t)=N_0e^{-\lambda t}
$$

This equation forms the basis of carbon dating.

---

The Meaning of the Constant C

When solving differential equations we repeatedly encounter constants.

For example:

$$
y=Ae^{3t}
$$

The constant

$$
A
$$

depends on the initial condition.

Suppose

$$
y(0)=5
$$

Then

$$
5=Ae^0
$$

Therefore

$$
A=5
$$

The solution becomes

$$
y=5e^{3t}
$$

Initial conditions transform a family of solutions into one unique solution.

---

A Useful Interpretation

Notice the structure

$$
\frac{dy}{dt}=ky
$$

appears repeatedly.

This equation says:

The rate of change is proportional to the current amount.

If

$$
k>0
$$

we obtain exponential growth.

If

$$
k<0
$$

we obtain exponential decay.

This single equation explains:

• Population growth
• Interest accumulation
• Epidemics (early stages)
• Radioactive decay
• Drug elimination
• Customer growth models
• Inventory replenishment models

---

The Big Idea

Separation of Variables allows us to transform a differential equation into an ordinary integration problem.

The process is:

1. Separate variables.
2. Integrate both sides.
3. Solve for the unknown function.
4. Use initial conditions to find constants.

Many important models in science and statistics come from this simple technique.

---

Looking Ahead

In Lesson 3 we will study:

First-Order Linear Differential Equations

We will learn how to solve equations that cannot be solved using simple separation of variables.

This introduces one of the most important tools in differential equations:

The Integrating Factor Method.

---

Key Takeaways

• Separation of Variables is one of the simplest methods for solving ODEs.
• Exponential growth follows

$$
\frac{dy}{dt}=ky
$$

with

$$
k>0
$$

• Exponential decay follows

$$
\frac{dy}{dt}=ky
$$

with

$$
k<0
$$

• Population growth and continuous compound interest share the same mathematical structure.
• Initial conditions determine a unique solution.
• Many real-world systems are governed by exponential growth or decay.