First Order Linear Differential Equation

Q: What makes the equation 'linear'?

An equation is linear if the dependent variable ($y$) and its derivatives all appear with a power of one. This means you won't see terms like $y^2$, $y^3$, or the variable multiplied by its own derivative.

Q: Why is the integrating factor always 'e' raised to a power?

It comes from the special property of the number $e$. The derivative of $e^{f(x)}$ is related to itself, which is exactly what's needed to reverse the product rule and simplify the left side of our differential equation.

Q: Do I always need to find the constant C?

For a 'general solution,' yes, you always leave $+C$ in the answer. If you are given an 'initial condition' (like the value of $y$ at a specific $x$), you use that information to find a specific number for $C$.

Q: What happens if I can't solve one of the integrals?

If the integral of $P(x)$ or $I(x)Q(x)$ is too difficult or impossible to solve with standard methods, then the differential equation can't be solved by hand this way. In those cases, mathematicians use computers to find approximate numerical solutions.

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Ever wondered how mathematicians predict population growth, track the cooling of a hot drink, or model the charge in a circuit? They use special equations that describe change. This guide will introduce you to one of the most fundamental types: first-order linear differential equations.

What Is a First-Order Linear Differential Equation?

A first-order linear differential equation is an equation that connects a function's value to its own rate of change in a simple, non-complex way. Think of it as a rule that says, "The speed at which this thing is changing is related to how big it is right now, plus some other external factors."

Let's break down that mouthful of a name:

Differential Equation: This is an equation that involves a function (like $y$ ) and its derivatives (like $\frac{d y}{d x}$ ). A derivative is just a fancy term for the instantaneous rate of change, or the "speed" at which the function's value is changing.
First-Order: This means we are only dealing with the first derivative ( $\frac{d y}{d x}$ ). We're not looking at acceleration or other higher-level rates of change. It’s all about the immediate speed.
Linear: This is a crucial word. It means the unknown function $y$ and its derivative $\frac{d y}{d x}$ only appear with a power of one. You won't see terms like $y^{2}$ , $(\frac{d y}{d x})^{3}$ , or $y \cdot \frac{d y}{d x}$ . This keeps the relationships simple and "straight-line" like, which makes them much easier to solve.

The standard form for these equations looks like this:

\frac{d y}{d x} + P (x) y = Q (x)

In this formula:

$y$ is the function we want to find (e.g., the temperature of your soup).
$x$ is the variable the function depends on (e.g., time).
$\frac{d y}{d x}$ is the rate of change of $y$ with respect to $x$ .
$P (x)$ and $Q (x)$ are functions of $x$ . They can be simple numbers (like $5$ ) or more complex expressions (like $3 x^{2}$ ). $P (x)$ tells us how the current amount $y$ affects the rate of change, and $Q (x)$ represents an external influence.

Why Are These Equations So Important?

These equations might seem abstract, but they are incredibly powerful tools for describing the world around us. So many real-life processes involve a quantity whose rate of change depends on its current size.

Here are just a few examples:

Compound Interest: The rate at which your money grows in a bank account ( $\frac{d M}{d t}$ ) is directly proportional to the amount of money ( $M$ ) you currently have.
Population Growth: In a simple model, the rate of population growth is proportional to the current population size. More people lead to more babies!
Newton's Law of Cooling: The rate at which a hot object cools down is proportional to the difference between its temperature and the surrounding room's temperature.
Radioactive Decay: The rate at which a radioactive substance decays is proportional to the amount of the substance present.

This table shows how we can frame these scenarios using the language of differential equations:

Real-World Scenario	The Changing Quantity ( $y$ )	What It Depends On ( $x$ )	What the Equation Describes
A cooling cup of tea	Temperature of the tea	Time	How fast the temperature drops.
A savings account	Account balance	Time	How quickly the balance grows.
A leaky water tank	Volume of water	Time	How fast the water level falls.

By solving these equations, scientists, engineers, and economists can make predictions, design systems, and understand the dynamics of the world.

What Is the Secret Weapon for Solving Them?

Solving these equations requires a clever trick. We can't just rearrange the equation to get $y$ by itself because $y$ and its derivative $\frac{d y}{d x}$ are mixed together. The secret weapon we use is called the integrating factor.

Think of the integrating factor as a special multiplier that transforms our difficult equation into one that's surprisingly easy to solve. When we multiply our entire equation by this factor, the left side magically turns into the result of a calculus rule called the "Product Rule" in reverse.

The formula for this magic multiplier, which we'll call $I (x)$ , is:

I (x) = e^{\int P (x) d x}

Let's quickly decode this formula:

The number $e$ is a special mathematical constant, approximately $2.718$ . It's often called Euler's number and is fundamental to describing growth and change.
The $\int P (x) d x$ part is an integral. For our purposes, think of integration as the opposite of differentiation. If differentiating tells you the speed, integrating takes the speed and tells you the original distance. So, $\int P (x) d x$ means "find the function whose derivative is $P (x)$ ."

Once we find this integrating factor, $I (x)$ , we'll see how it simplifies the whole process.

How Do You Solve a First-Order Linear Differential Equation?

Solving these equations follows a reliable, step-by-step recipe. As long as you follow the steps carefully, you'll be able to solve any equation in this form.

Write in Standard Form: First, make sure your equation looks exactly like this: $\frac{d y}{d x} + P (x) y = Q (x)$ . You might need to add, subtract, multiply, or divide to get it into this format.
Find the Integrating Factor ( $I (x)$ ): Identify the $P (x)$ term from your standard form equation. Then, calculate the integrating factor using the formula $I (x) = e^{\int P (x) d x}$ .
Multiply Everything: Take the integrating factor you just found and multiply every single term in your standard form equation by it. This gives you: $I (x) \frac{d y}{d x} + I (x) P (x) y = I (x) Q (x)$ .
Simplify and Integrate: Here's the magic part. The left side of the equation, $I (x) \frac{d y}{d x} + I (x) P (x) y$ , is now guaranteed to be the derivative of the product $I (x) y$ . So, you can rewrite the equation as $\frac{d}{d x} (I (x) y) = I (x) Q (x)$ . Now, integrate both sides with respect to $x$ . This will cancel out the derivative on the left, leaving you with: $I (x) y = \int I (x) Q (x) d x$ .
Solve for $y$ : To get your final answer, just divide both sides by the integrating factor, $I (x)$ . Don't forget to include the constant of integration, $+ C$ , that appears when you perform the integration in step 4!

Worked Example 1: A Simple Case

Example 1

Solve the differential equation $\frac{d y}{d x} + 3 y = 9$ .

Step 1: Standard Form
The equation is already in the standard form $\frac{d y}{d x} + P (x) y = Q (x)$ . By comparing, we can see that $P (x) = 3$ and $Q (x) = 9$ .

Step 2: Find the Integrating Factor
We use the formula $I (x) = e^{\int P (x) d x}$ .
First, we find the integral of $P (x)$ : $\int 3 d x = 3 x$ . (We don't need the constant of integration here.)
So, our integrating factor is $I (x) = e^{3 x}$ .

Step 3: Multiply Everything
We multiply our original equation by $e^{3 x}$ :
$e^{3 x} \frac{d y}{d x} + e^{3 x} (3 y) = e^{3 x} (9)$

Step 4: Simplify and Integrate
The left side is now the derivative of $I (x) y$ , which is $\frac{d}{d x} (e^{3 x} y)$ . So our equation becomes:
$\frac{d}{d x} (e^{3 x} y) = 9 e^{3 x}$
Now, we integrate both sides:
$\int \frac{d}{d x} (e^{3 x} y) d x = \int 9 e^{3 x} d x$
$e^{3 x} y = 9 (\frac{1}{3} e^{3 x}) + C$
$e^{3 x} y = 3 e^{3 x} + C$

Step 5: Solve for $y$
To isolate $y$ , we divide everything by $e^{3 x}$ :
$y = \frac{3 e^{3 x}}{e^{3 x}} + \frac{C}{e^{3 x}}$
Our final general solution is:
$y = 3 + C e^{- 3 x}$

Worked Example 2: When P(x) Isn't a Constant

Example 2

Solve the differential equation $x \frac{d y}{d x} - 2 y = x^{3}$ , assuming $x > 0$ .

Step 1: Standard Form
This equation is not in standard form because of the $x$ in front of $\frac{d y}{d x}$ . We must divide the entire equation by $x$ :
$\frac{d y}{d x} - \frac{2}{x} y = x^{2}$
Now it's in standard form. We can see that $P (x) = - \frac{2}{x}$ and $Q (x) = x^{2}$ .

Step 2: Find the Integrating Factor
We calculate $I (x) = e^{\int P (x) d x}$ .
First, the integral: $\int - \frac{2}{x} d x = - 2 \int \frac{1}{x} d x = - 2 \ln (x)$ . (We use $\ln (x)$ since we are told $x > 0$ .)
Using a logarithm rule, we can write this as $\ln (x^{- 2})$ .
So, our integrating factor is $I (x) = e^{\ln (x^{- 2})} = x^{- 2} = \frac{1}{x^{2}}$ .

Step 3: Multiply Everything
We multiply our standard form equation by $\frac{1}{x^{2}}$ :
$\frac{1}{x^{2}} \frac{d y}{d x} - \frac{1}{x^{2}} \frac{2}{x} y = \frac{1}{x^{2}} x^{2}$
$\frac{1}{x^{2}} \frac{d y}{d x} - \frac{2}{x^{3}} y = 1$

Step 4: Simplify and Integrate
The left side is now equal to $\frac{d}{d x} (\frac{1}{x^{2}} y)$ . So:
$\frac{d}{d x} (\frac{1}{x^{2}} y) = 1$
Integrate both sides:
$\int \frac{d}{d x} (\frac{1}{x^{2}} y) d x = \int 1 d x$
$\frac{1}{x^{2}} y = x + C$

Step 5: Solve for $y$
Multiply by $x^{2}$ to get $y$ by itself:
$y = x^{2} (x + C)$
Our final general solution is:
$y = x^{3} + C x^{2}$

Worked Example 3: Finding a Specific Solution with an Initial Value

Sometimes, we are given a piece of information called an "initial condition," which tells us the value of $y$ at a specific point $x$ . This allows us to find the exact value of the constant $C$ and get a single, specific solution.

Example 3

Solve $\frac{d y}{d t} + y = 5$ given the initial condition $y (0) = 1$ . (Note: We use $t$ for time instead of $x$ , but the method is identical).

Steps 1-2: Standard Form and Integrating Factor
The equation is in standard form with $P (t) = 1$ and $Q (t) = 5$ .
The integrating factor is $I (t) = e^{\int 1 d t} = e^{t}$ .

Steps 3-4: Multiply and Integrate
Multiply by $e^{t}$ : $e^{t} \frac{d y}{d t} + e^{t} y = 5 e^{t}$ .
This simplifies to $\frac{d}{d t} (e^{t} y) = 5 e^{t}$ .
Integrating both sides gives: $e^{t} y = \int 5 e^{t} d t = 5 e^{t} + C$ .

Step 5: Solve for $y$ (General Solution)
Divide by $e^{t}$ :
$y = 5 + \frac{C}{e^{t}} = 5 + C e^{- t}$
This is our general solution.

Final Step: Use the Initial Condition
Now we use the fact that $y (0) = 1$ . This means when $t = 0$ , $y = 1$ . We plug these values into our general solution:
$1 = 5 + C e^{- 0}$
Since $e^{0} = 1$ , this becomes:
$1 = 5 + C (1)$
$1 = 5 + C$
Solving for $C$ , we find $C = 1 - 5 = - 4$ .
Now we substitute this value of $C$ back into our general solution to get the specific solution:
$y = 5 - 4 e^{- t}$

What Are Some Common Mistakes to Avoid?

This method is very powerful, but small mistakes can lead to the wrong answer. Be on the lookout for these common pitfalls:

Forgetting Standard Form: Before you identify $P (x)$ and $Q (x)$ , you must ensure the equation is in the form $\frac{d y}{d x} + P (x) y = Q (x)$ . This often means dividing by a term in front of $\frac{d y}{d x}$ .
Errors in Integration: The most common place for errors is when calculating the two integrals: $\int P (x) d x$ for the integrating factor and $\int I (x) Q (x) d x$ for the solution. Double-check your integration rules!
Forgetting to Multiply $Q (x)$ : When you multiply by the integrating factor $I (x)$ , remember to multiply the right side of the equation, $Q (x)$ , as well. It's easy to forget.
Forgetting the Constant of Integration $+ C$ : Every time you perform an indefinite integration, you must add a constant $+ C$ . Forgetting it means you will only find one possible solution, not the whole family of them.
Logarithm and Exponent Mistakes: When $P (x)$ involves terms like $\frac{1}{x}$ , you will encounter natural logarithms ( $\ln$ ) in your integrating factor. Be careful with your exponent and log rules, like $e^{\ln (a)} = a$ and $b \ln (a) = \ln (a^{b})$ .

Quick Summary and Key Formulas

Here's a quick reference for everything we've covered.

A first-order linear differential equation describes a relationship between a function $y$ and its first derivative $\frac{d y}{d x}$ .

Standard Form:

\frac{d y}{d x} + P (x) y = Q (x)

The 5-Step Solution Method:

Get the equation into Standard Form.
Find the Integrating Factor: $I (x) = e^{\int P (x) d x}$ .
Multiply the standard form equation by $I (x)$ .
Rewrite and integrate: $\frac{d}{d x} (I (x) y) = I (x) Q (x) ⟹ I (x) y = \int I (x) Q (x) d x$ .
Solve for $y$ : $y = \frac{1}{I (x)} (\int I (x) Q (x) d x)$ .

Remember to always include the constant of integration $+ C$ and use any initial values to find its specific value.

Frequently Asked Questions

What does 'first-order' actually mean?

'First-order' refers to the highest derivative in the equation. Since we only have the first derivative ( $\frac{d y}{d x}$ ) and not the second derivative or higher, it's called a first-order equation.

What makes the equation 'linear'?

An equation is linear if the dependent variable ( $y$ ) and its derivatives all appear with a power of one. This means you won't see terms like $y^{2}$ , $y^{3}$ , or the variable multiplied by its own derivative.

Why is the integrating factor always 'e' raised to a power?

It comes from the special property of the number $e$ . The derivative of $e^{f (x)}$ is related to itself, which is exactly what's needed to reverse the product rule and simplify the left side of our differential equation.

Do I always need to find the constant C?

For a 'general solution,' yes, you always leave $+ C$ in the answer. If you are given an 'initial condition' (like the value of $y$ at a specific $x$ ), you use that information to find a specific number for $C$ .

Is this topic part of Algebra or Calculus?

It's a key topic in a branch of mathematics called Differential Equations, which is typically studied after Algebra and introductory Calculus. It uses skills from both subjects, especially integration from Calculus.

What happens if I can't solve one of the integrals?

If the integral of $P (x)$ or $I (x) Q (x)$ is too difficult or impossible to solve with standard methods, then the differential equation can't be solved by hand this way. In those cases, mathematicians use computers to find approximate numerical solutions.

Can a calculator solve these equations?

Many advanced graphing calculators and online tools like WolframAlpha can solve differential equations automatically. However, it's crucial to learn the manual method to understand what the calculator is actually doing.