Linear First-Order Equations

Welcome, students! 👋 In this lesson, you will learn one of the most important types of differential equations: the linear first-order equation. These equations appear in science, engineering, economics, and everyday models such as cooling, population change, and mixing liquids. By the end of this lesson, you should be able to recognize a linear first-order equation, understand the main terminology, and solve it using the integrating factor method.

What is a linear first-order equation?

A first-order differential equation contains only the first derivative of the unknown function. That means the highest derivative present is $\frac{dy}{dx}$. A linear first-order differential equation is one that can be written in the form

$$\frac{dy}{dx} + P(x)y = Q(x)$$

where $P(x)$ and $Q(x)$ are known functions of $x$.

This equation is called linear because the unknown function $y$ and its derivative $\frac{dy}{dx}$ appear only to the first power, and they are not multiplied together. For example, $y^2$, $\sin(y)$, and $y\frac{dy}{dx}$ do not fit the linear form.

A key idea is that the equation describes how a quantity changes with respect to another quantity. For example, $y(x)$ might represent temperature, amount of medicine in the bloodstream, height of water in a tank, or the concentration of salt in a solution. 🌍

Main terminology

• $y$ is the dependent variable.
• $x$ is the independent variable.
• $\frac{dy}{dx}$ is the derivative of $y$ with respect to $x$.
• $P(x)$ and $Q(x)$ are known coefficient functions.
• A solution is a function $y(x)$ that makes the equation true.

Understanding this standard form is the first big step because many differential equations can be rearranged into it.

Why linearity matters

The word linear matters because it gives us a reliable method for solving the equation. Nonlinear equations can behave in complicated ways, but linear first-order equations have a structured solution process.

If we have

$$\frac{dy}{dx} + P(x)y = Q(x),$$

then we can use an integrating factor to turn the left side into a derivative of a product. This makes the equation much easier to solve.

The reason this works is based on the product rule from calculus:

$$\frac{d}{dx}[\mu(x)y(x)] = \mu(x)\frac{dy}{dx} + \mu'(x)y(x).$$

If we choose $\mu(x)$ carefully, the extra term $\mu'(x)y(x)$ will match the $P(x)y$ part of the differential equation.

This connection between differential equations and ordinary calculus is one of the main ideas in the topic First-Order Equations II. It shows how a difficult equation can be transformed into a simpler one by using algebra and derivative rules together.

The integrating factor method

To solve

$$\frac{dy}{dx} + P(x)y = Q(x),$$

we multiply both sides by an integrating factor $\mu(x)$, which is defined by

$$\mu(x)=e^{\int P(x)\,dx}.$$

After multiplying through, the equation becomes

$$\mu(x)\frac{dy}{dx} + \mu(x)P(x)y = \mu(x)Q(x).$$

Because of the way $\mu(x)$ is chosen, the left side becomes

$$\frac{d}{dx}[\mu(x)y] = \mu(x)Q(x).$$

Now we integrate both sides:

$$\mu(x)y = \int \mu(x)Q(x)\,dx + C,$$

and then solve for $y$:

$$y = \frac{1}{\mu(x)}\left(\int \mu(x)Q(x)\,dx + C\right).$$

This is the general solution.

Why the integrating factor works

students, think of the integrating factor as a “helper function” that turns two separate terms into one combined derivative. It is a clever way of using the product rule in reverse. 🧠

Without the integrating factor, the equation might not be easy to integrate directly. With it, the equation becomes manageable because the left side collapses into a single derivative.

Example 1: Solving a simple linear equation

Let’s solve

$$\frac{dy}{dx} + 2y = 6.$$

Here, $P(x)=2$ and $Q(x)=6$.

Step 1: Find the integrating factor

$$\mu(x)=e^{\int 2\,dx}=e^{2x}.$$

Step 2: Multiply the entire equation by $e^{2x}$

$$e^{2x}\frac{dy}{dx}+2e^{2x}y=6e^{2x}.$$

The left side is the derivative of a product:

$$\frac{d}{dx}[e^{2x}y]=6e^{2x}.$$

Step 3: Integrate both sides

$$e^{2x}y=\int 6e^{2x}\,dx+C.$$

Since $\int 6e^{2x}\,dx=3e^{2x}$, we get

$$e^{2x}y=3e^{2x}+C.$$

Step 4: Solve for $y$

$$y=3+Ce^{-2x}.$$

This solution has two parts:

• $3$, which is a particular solution
• $Ce^{-2x}$, which is the homogeneous part

If an initial condition were given, such as $y(0)=5$, we could find $C$.

Example 2: A variable coefficient equation

Now consider

$$\frac{dy}{dx}+\frac{1}{x}y=x,$$

where $x>0$.

Here, $P(x)=\frac{1}{x}$ and $Q(x)=x$.

Step 1: Find the integrating factor

$$\mu(x)=e^{\int \frac{1}{x}\,dx}=e^{\ln x}=x.$$

Step 2: Multiply through by $x$

$$x\frac{dy}{dx}+y=x^2.$$

The left side is now

$$\frac{d}{dx}[xy]=x^2.$$

Step 3: Integrate

$$xy=\int x^2\,dx+C=\frac{x^3}{3}+C.$$

Step 4: Solve for $y$

$$y=\frac{x^2}{3}+\frac{C}{x}.$$

This example shows that the integrating factor may be a simple function like $x$ or a more complicated one, depending on $P(x)$.

How to recognize linear first-order equations

A common challenge is deciding whether an equation is linear. Use these checks:

1. The highest derivative should be $\frac{dy}{dx}$.
2. The function $y$ and its derivative should appear only to the first power.
3. There should be no products like $y\frac{dy}{dx}$.
4. There should be no powers like $y^2$.
5. The equation should be rearrangeable into $\frac{dy}{dx}+P(x)y=Q(x)$.

Examples of linear equations:

$$\frac{dy}{dx}+3y=\sin x$$

$$x\frac{dy}{dx}+y=x^2\quad (x\neq 0)$$

$$\frac{dy}{dx}-\frac{2}{x}y=e^x$$

Examples of non-linear equations:

$$\frac{dy}{dx}+y^2=x$$

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

$$\frac{dy}{dx}+\sin(y)=x$$

Recognizing the form quickly helps you choose the correct method. ⚙️

Connection to First-Order Equations II and real-world models

Linear first-order equations are a major part of the broader topic First-Order Equations II because they introduce a standard technique for solving models where change depends on both the current state and an external input.

A common real-world interpretation is growth or decay with forcing. The term $P(x)y$ represents a process that depends on the current amount $y$, while $Q(x)$ represents an outside influence. For example:

• In a cooling model, $P(x)y$ can represent heat loss proportional to the temperature difference.
• In a tank problem, $Q(x)$ may represent salt entering the tank from an outside source.
• In a population model, $Q(x)$ could represent immigration or harvesting.

These equations often model situations where the rate of change is not just caused by the quantity itself, but also by something acting from outside. That is why linear first-order equations are so useful in applied mathematics.

Solving with an initial condition

Many real problems include an initial condition, such as $y(x_0)=y_0$. This gives one specific solution instead of a whole family.

For example, suppose

$$\frac{dy}{dx}+2y=6,\qquad y(0)=5.$$

From before, the general solution is

$$y=3+Ce^{-2x}.$$

Now use $y(0)=5$:

$$5=3+C.$$

So $C=2$, and the particular solution is

$$y=3+2e^{-2x}.$$

An initial condition turns an infinite set of possible solutions into one exact answer.

Common mistakes to avoid

students, here are mistakes students often make:

• Forgetting to check whether the equation is in linear form.
• Using the wrong integrating factor.
• Forgetting to multiply both sides by the integrating factor.
• Missing the product rule when rewriting the left-hand side.
• Ignoring the constant of integration $C$.
• Not solving for $y$ at the end.

A good habit is to write each step clearly and check that your final answer fits the original equation.

Conclusion

Linear first-order equations are a central topic in differential equations because they connect algebra, calculus, and modeling in a powerful way. The standard form

$$\frac{dy}{dx}+P(x)y=Q(x)$$

lets us identify the equation quickly, and the integrating factor method gives a systematic way to solve it. These equations are especially important in applications where a changing quantity is influenced by both its current value and an outside effect. By mastering this lesson, students, you are building a strong foundation for the rest of First-Order Equations II. ✅

Study Notes

• A linear first-order differential equation has the form $\frac{dy}{dx}+P(x)y=Q(x)$.
• It is called linear because $y$ and $\frac{dy}{dx}$ appear only to the first power and are not multiplied together.
• The integrating factor is $\mu(x)=e^{\int P(x)\,dx}$.
• Multiplying the equation by $\mu(x)$ turns the left side into $\frac{d}{dx}[\mu(x)y]$.
• After integrating, the general solution is $y=\frac{1}{\mu(x)}\left(\int \mu(x)Q(x)\,dx+C\right)$.
• Initial conditions like $y(x_0)=y_0$ let you find the constant $C$.
• Linear first-order equations are widely used in models of cooling, mixing, population change, and more.
• Recognizing linear form is the key step before solving.