algebra911.com

Linear Differential Equation

Q: Why is it called a 'differential' equation?

It's named after the 'differentials' (like $dx$ and $dy$) or 'derivatives' ($ \frac{dy}{dx} $) it contains. A derivative is a way to measure how a function changes, so a differential equation is fundamentally an equation about rates of change.

Q: Can a differential equation have more than one derivative?

Absolutely! An equation involving the second derivative ($ \frac{d^2y}{dx^2} $) is called a second-order differential equation. They are used to model things like vibrations and waves, but the solving methods are more advanced.

Download as PDF

Ever wondered how scientists model population growth or how a hot cup of coffee cools down? The secret lies in a special type of math equation that describes change. Welcome to the world of linear differential equations, a powerful tool for understanding the world around us.

Linear Differential Equation — an original Algebra911 reference diagram defining linear differential equation with its key formula and a worked example. — An Introduction to Linear Differential Equations

What Is a Linear Differential Equation?

A linear differential equation is an equation that connects a function with its derivatives in a linear way. That might sound like a mouthful, but let's break it down. You're already familiar with equations like $y = 2 x + 1$ . A differential equation is similar, but it includes not just a variable $y$ , but also its rate of change.

Think about it this way:

Equation: It has an equals sign, connecting two mathematical expressions.
Differential: This part means it involves a special kind of rate called a derivative. A derivative is just a precise way of talking about slope or how fast something is changing at any given moment. If you think of your position on a road trip, your derivative is your speed on the speedometer.
Linear: This is a crucial word. It means the function (let's call it $y$ ) and its derivatives (like $y^{'}$ or $\frac{d y}{d x}$ ) are not squared, cubed, or multiplied by each other. They behave like variables in a simple line equation, $y = m x + b$ , with no tricky exponents.

So, a linear differential equation is a relationship that describes how the rate of change of a quantity is connected to the quantity itself, in a straightforward, non-complicated way.

Understanding the Building Blocks: What's a Derivative?

The heart of a differential equation is the derivative. The most common symbol for a derivative is $\frac{d y}{d x}$ . This notation looks like a fraction, and in a way, it is. It means "the change in $y$ for a tiny change in $x$ ". In simpler terms, it's the instantaneous rate of change.

Let's use an example you already know: the equation of a line, $y = 3 x + 5$ . What is the rate of change here? For every 1 unit you move to the right on the x-axis, the y-value goes up by 3. The slope is constant, and it's always 3. So, we can write a very simple differential equation:

\frac{d y}{d x} = 3

This equation tells us that the rate of change of $y$ with respect to $x$ is always 3. The original function, $y = 3 x + 5$ , is a solution to this differential equation. (So is $y = 3 x + 10$ and $y = 3 x - 2$ ! We'll see why later.)

For a curve, like $y = x^{2}$ , the slope isn't constant. It's steep in some places and flat in others. The derivative tells you the exact slope at any single point. A differential equation uses this changing slope to define the function itself.

What Makes a Differential Equation 'Linear'?

The term "linear" is very specific. It means that the variable representing our unknown function (we'll call it $y$ ) and all of its derivatives ( $\frac{d y}{d x}$ , $\frac{d^{2} y}{d x^{2}}$ , etc.) only appear with an exponent of 1. They aren't multiplied together or trapped inside another function like $\sin (y)$ or $e^{y}$ .

The standard form for a first-order (meaning it only has the first derivative) linear differential equation is:

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

Here's what each part means:

$\frac{d y}{d x}$ is the first derivative of the unknown function $y$ .
$P (x)$ is a function that only involves $x$ . It's the coefficient of $y$ .
$y$ is the unknown function we are trying to find.
$Q (x)$ is another function that only involves $x$ . It's the term on the other side of the equation.

Let's see what's linear and what isn't with a table:

Example Equation	Is it Linear?	Reason
$\frac{d y}{d x} + 5 y = x^{2}$	Yes	It perfectly fits the form $\frac{d y}{d x} + P (x) y = Q (x)$ , with $P (x) = 5$ and $Q (x) = x^{2}$ .
$x \frac{d y}{d x} - 2 y = 0$	Yes	We can divide by $x$ to get $\frac{d y}{d x} - \frac{2}{x} y = 0$ . Here, $P (x) = - \frac{2}{x}$ and $Q (x) = 0$ .
$\frac{d y}{d x} + y^{2} = 5$	No	The $y$ term is squared ( $y^{2}$ ), which makes it non-linear.
$y \frac{d y}{d x} + 2 x = 0$	No	The function $y$ is multiplied by its own derivative $\frac{d y}{d x}$ , which is a non-linear combination.

For the rest of this lesson, we will focus only on solving the linear ones!

How Do You Solve First-Order Linear Equations?

Solving a linear differential equation means finding the function $y$ that makes the equation true. The most reliable method for first-order linear equations is called the Integrating Factor Method. It's like having a secret key that turns a complicated equation into something easy to solve. The goal is to manipulate the equation so we can just integrate both sides.

The process involves a special function called the integrating factor, usually denoted $I (x)$ . We calculate it and multiply our entire differential equation by it. This magically transforms the left side of the equation into the result of a derivative product rule, which we can then easily reverse by integrating.

Here is the step-by-step recipe:

Standard Form: Make sure your equation is in the standard form: $\frac{d y}{d x} + P (x) y = Q (x)$ . You must isolate $\frac{d y}{d x}$ .
Find P(x): Identify the function $P (x)$ that is being multiplied by $y$ .
Calculate Integrating Factor $I (x)$ : The formula for the integrating factor is:
I(x) = e^{\int P(x) dx}
This means you find the integral (or antiderivative) of $P (x)$ , and then make that the exponent of $e$ . Don't worry about the constant of integration ( $+ C$ ) for this step.
Multiply: Multiply every term in the standard form equation by your integrating factor $I (x)$ .
Simplify and Integrate: The left side of your new equation will always simplify to $\frac{d}{d x} [I (x) \cdot y]$ . So your equation becomes $\frac{d}{d x} [I (x) \cdot y] = I (x) Q (x)$ . Now you can integrate both sides with respect to $x$ .
Solve for y: After integrating, you'll have $I (x) \cdot y = \int I (x) Q (x) d x$ . The final step is to divide by $I (x)$ to get $y$ by itself. This time, you absolutely must include the constant of integration, $C$ .

Let's see it in action.

Example 1

Solve the differential equation $\frac{d y}{d x} + 2 y = 6$ .

Step 1: Standard Form. The equation is already in standard form $\frac{d y}{d x} + P (x) y = Q (x)$ .

Step 2: Find P(x). Comparing our equation to the standard form, we see that $P (x) = 2$ and $Q (x) = 6$ .

Step 3: Calculate Integrating Factor. We use the formula $I (x) = e^{\int P (x) d x}$ .
$I (x) = e^{\int 2 d x} = e^{2 x}$ Our integrating factor is $e^{2 x}$ .

Step 4: Multiply. We multiply the entire original equation by $e^{2 x}$ :
$e^{2 x} (\frac{d y}{d x} + 2 y) = e^{2 x} (6)$ $e^{2 x} \frac{d y}{d x} + 2 e^{2 x} y = 6 e^{2 x}$

Step 5: Simplify and Integrate. The left side is now the derivative of $I (x) \cdot y$ , which is $\frac{d}{d x} (e^{2 x} y)$ . So we have:
$\frac{d}{d x} (e^{2 x} y) = 6 e^{2 x}$ Now, we integrate both sides with respect to $x$ :
$\int \frac{d}{d x} (e^{2 x} y) d x = \int 6 e^{2 x} d x$ $e^{2 x} y = 3 e^{2 x} + C$ (Remember to include the constant of integration $C$ !)

Step 6: Solve for y. To get $y$ by itself, we divide everything by $e^{2 x}$ :
$y = \frac{3 e^{2 x}}{e^{2 x}} + \frac{C}{e^{2 x}}$ $y = 3 + C e^{- 2 x}$ This is the general solution to the differential equation.

How to Handle a More Complex Example

The previous example was straightforward because $P (x)$ and $Q (x)$ were constants. Let's try one where they are functions of $x$ .

Example 2

Solve the differential equation $x \frac{d y}{d x} - 3 y = x^{5}$ , given the initial condition $y (1) = 0$ .

Step 1: Standard Form. The 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 $x$ :
$\frac{d y}{d x} - \frac{3}{x} y = x^{4}$ Now it's in standard form.

Step 2: Find P(x). From the standard form, we can see that $P (x) = - \frac{3}{x}$ and $Q (x) = x^{4}$ .

Step 3: Calculate Integrating Factor. We need to find $\int P (x) d x$ :
$\int - \frac{3}{x} d x = - 3 \int \frac{1}{x} d x = - 3 \ln (x)$ We can use logarithm properties to write this as $\ln (x^{- 3})$ . Now we find $I (x)$ :
$I (x) = e^{\ln (x^{- 3})} = x^{- 3} = \frac{1}{x^{3}}$ Our integrating factor is $\frac{1}{x^{3}}$ .

Step 4: Multiply. Multiply the standard form equation by $\frac{1}{x^{3}}$ :
$\frac{1}{x^{3}} (\frac{d y}{d x} - \frac{3}{x} y) = \frac{1}{x^{3}} (x^{4})$ $\frac{1}{x^{3}} \frac{d y}{d x} - \frac{3}{x^{4}} y = x$

Step 5: Simplify and Integrate. The left side is $\frac{d}{d x} (\frac{1}{x^{3}} y)$ . So:
$\frac{d}{d x} (\frac{1}{x^{3}} y) = x$ Integrate both sides:
$\int \frac{d}{d x} (\frac{1}{x^{3}} y) d x = \int x d x$ $\frac{1}{x^{3}} y = \frac{1}{2} x^{2} + C$

Step 6: Solve for y. Multiply by $x^{3}$ to isolate $y$ :
$y = x^{3} (\frac{1}{2} x^{2} + C)$ $y = \frac{1}{2} x^{5} + C x^{3}$ This is the general solution. Now we use the initial condition $y (1) = 0$ to find the specific value of $C$ . This means when $x = 1$ , $y = 0$ .
$0 = \frac{1}{2} (1)^{5} + C (1)^{3}$ $0 = \frac{1}{2} + C$ $C = - \frac{1}{2}$ Substituting this value of $C$ back into our general solution gives the final answer:
$y = \frac{1}{2} x^{5} - \frac{1}{2} x^{3}$

Key formulas for linear differential equation by Algebra911.

Where Are These Used? A Real-World Example

Differential equations are essential in science and engineering. One classic example is Newton's Law of Cooling, which describes how an object's temperature changes over time.

The law states that the rate of change of an object's temperature is proportional to the difference between its own temperature and the temperature of its surroundings. We can write this as a differential equation:

\frac{d T}{d t} = - k (T - T_{a})

Where $T$ is the object's temperature, $t$ is time, $T_{a}$ is the ambient (surrounding) temperature, and $k$ is a positive constant that depends on the object's physical properties.

Example 3

A pie is taken out of an oven at $180^{\circ} F$ and placed in a room where the temperature is $70^{\circ} F$ . The differential equation modeling this is $\frac{d T}{d t} = - k (T - 70)$ . Find the temperature of the pie at any time $t$ .

Step 1: Get into Linear Form. First, distribute the $- k$ :
$\frac{d T}{d t} = - k T + 70 k$ Now, move the $- k T$ term to the left side to match the standard form $\frac{d y}{d x} + P (x) y = Q (x)$ . Our variables are $T$ and $t$ instead of $y$ and $x$ .
$\frac{d T}{d t} + k T = 70 k$ This is a linear differential equation with $P (t) = k$ and $Q (t) = 70 k$ .

Step 2: Find the Integrating Factor.
$I (t) = e^{\int k d t} = e^{k t}$

Step 3: Multiply and Solve. Multiply the linear form by $e^{k t}$ :
$e^{k t} \frac{d T}{d t} + k e^{k t} T = 70 k e^{k t}$ The left side becomes $\frac{d}{d t} (e^{k t} T)$ :
$\frac{d}{d t} (e^{k t} T) = 70 k e^{k t}$ Integrate both sides with respect to $t$ :
$e^{k t} T = \int 70 k e^{k t} d t = 70 e^{k t} + C$

Step 4: Isolate T. Divide by $e^{k t}$ :
$T (t) = 70 + C e^{- k t}$ This is the general solution. We can use the initial condition—at $t = 0$ , the temperature was $180^{\circ} F$ —to find $C$ .
$180 = 70 + C e^{- k (0)}$ $180 = 70 + C (1)$ $C = 110$ So the specific solution for the pie's temperature at any time $t$ is:
$T (t) = 70 + 110 e^{- k t}$ To find the value of $k$ , we would need another data point, like the temperature after 10 minutes.

Common Mistakes to Avoid

Forgetting the Constant of Integration (C): This is the most common mistake. Every time you perform an indefinite integration, you must add $+ C$ . This constant is what gives you the general solution. Without it, you only have one of infinitely many possible answers.
Incorrectly Identifying P(x): Always make sure the equation is in standard form $\frac{d y}{d x} + P (x) y = Q (x)$ before you identify $P (x)$ . If you have $3 \frac{d y}{d x} + . . .$ , you must divide the whole equation by 3 first. The coefficient of $P (x)$ includes the sign, so if you have $\frac{d y}{d x} - 2 x y = . . .$ , then $P (x) = - 2 x$ .
Errors in Integration: The method relies on correctly integrating $P (x)$ to find the integrating factor and later integrating $I (x) Q (x)$ . Be careful with your integration rules, especially for logarithms and exponential functions.
Logarithm/Exponent Rule Mistakes: Remember that $e^{\ln (A)} = A$ and $m \ln (A) = \ln (A^{m})$ . These are crucial for simplifying the integrating factor after you integrate $P (x)$ .
Applying Initial Conditions Too Early: You must find the full general solution (with the $+ C$ ) before you plug in the initial conditions to solve for $C$ .

Quick Reference Guide

Here's a quick summary of the key steps for solving a first-order linear differential equation.

1. The Goal: Find the function $y (x)$ that solves an equation of the form $\frac{d y}{d x} + P (x) y = Q (x)$ .

2. The Method: Integrating Factor

Standard Form: Write the equation as $\frac{d y}{d x} + P (x) y = Q (x)$ .
Find P(x): Identify the coefficient of the $y$ term.
Find Integrating Factor $I (x)$ : Calculate $I (x) = e^{\int P (x) d x}$ .
Multiply: Multiply the standard form equation by $I (x)$ .
Recognize and Integrate: The equation becomes $\frac{d}{d x} (I (x) \cdot y) = I (x) Q (x)$ . Integrate both sides.
Solve for y: Isolate $y$ and don't forget the constant of integration, $C$ .

3. Key Formulas:

Standard Form:
$\frac{d y}{d x} + P (x) y = Q (x)$
Integrating Factor:
$I (x) = e^{\int P (x) d x}$
After Multiplying:
$\frac{d}{d x} (I (x) \cdot y) = I (x) Q (x)$

Frequently Asked Questions

Is this calculus? It seems complicated for my grade level.

Yes, this topic is part of calculus. This lesson is a gentle introduction to the ideas, connecting them to algebra concepts you already know like slope and equations. We focus on the 'why' and 'how' without getting lost in complex theory.

What does the constant 'C' in the solution mean?

The constant of integration, 'C', represents a whole family of possible solution curves. To find a single, specific solution, you need an 'initial condition'—a known point $(x, y)$ that your solution must pass through. This allows you to solve for the exact value of C.

Why is it called a 'differential' equation?

It's named after the 'differentials' (like $d x$ and $d y$ ) or 'derivatives' ( $\frac{d y}{d x}$ ) it contains. A derivative is a way to measure how a function changes, so a differential equation is fundamentally an equation about rates of change.

Can a differential equation have more than one derivative?

Absolutely! An equation involving the second derivative ( $\frac{d^{2} y}{d x^{2}}$ ) is called a second-order differential equation. They are used to model things like vibrations and waves, but the solving methods are more advanced.

Do I always have to use the integrating factor method?

No. For some simpler linear differential equations, a method called 'separation of variables' works and can be faster. However, the integrating factor method is more powerful because it can solve any first-order linear differential equation, which is why we focus on it.

What's the difference between P(x) and p(x)?

In mathematics, capitalization can sometimes be important, but in this context, $P (x)$ and $p (x)$ are just labels for a function of $x$ . Most textbooks use a capital $P (x)$ for the standard form, so it's good practice to stick with that to avoid confusion.

Are there non-linear differential equations?

Yes, many! Non-linear differential equations are those where the function or its derivatives are squared, multiplied together, or inside another function (like $\sin (y)$ ). They model more complex phenomena, like chaotic weather patterns, and are generally much harder to solve.