Standard Form of First-Order Linear DEs

What Makes a Differential Equation "First-Order Linear"?

Before we can standardize an equation, we need to be sure we're dealing with the correct type. A differential equation is called first-order if the highest derivative present is the first derivative (like $y^{\prime }$ or $\frac{dy}{dx}$).

It is called linear if the dependent variable ($y$) and its derivative ($y^{\prime }$) both appear only to the first power and are not part of other functions (like $\sin (y)$ or $e^y$). The functions multiplying $y$ and $y^{\prime }$ can be any functions of the independent variable, $x$.

To get a formal definition and see clear examples of what is and isn't a linear equation, please read the first part of the following resource.

A key point to note is that linearity depends on the dependent variable ($y$), not the independent variable ($x$). For instance, an equation like $y^{\prime }+(\cos x)y=x^2$ is linear, even though $\cos x$ and $x^2$ are not linear functions of $x$.

To give you a bird's-eye view, here is a flowchart that shows how first-order linear equations fit into the broader family of first-order differential equations. We will be focusing on the path leading to the "First Order Linear" box.

The Standard Form

Now that we can identify a first-order linear equation, our next step is to write it in a consistent, predictable format. This is known as the standard form.

The standard form for a first-order linear differential equation is:

or, using prime notation:

Here:

• The coefficient of the derivative $y^{\prime }$ is exactly 1.
• $p(x)$ is the function of $x$ that multiplies $y$.
• $q(x)$ is the function of $x$ on the other side of the equation.

Any first-order linear differential equation can be rearranged into this form. Just as standardizing a data format is crucial for processing it with a consistent algorithm, putting a differential equation into its standard form is the necessary first step before applying a general solution method.

The following reading explains this process and shows how to derive the standard form from the more general form $a(x)y^{\prime }+b(x)y=c(x)$.

Practice: Converting to Standard Form

The best way to get comfortable with this concept is through practice. The process usually involves simple algebraic manipulation to isolate $y^{\prime }$ with a coefficient of 1.

The following video provides several excellent examples. For each one, the presenter starts with an equation that is not in standard form, manipulates it, and then identifies $p(x)$ and $q(x)$.

After watching, let's break down the general steps:

1. Arrange the terms: Move all terms involving $y$ and $y^{\prime }$ to one side of the equation and all other terms to the other side.
2. Isolate the $y^{\prime }$ term: Ensure the term with the derivative stands alone on one side or is grouped with the $y$ term.
3. Normalize the equation: Divide the entire equation by the coefficient of $y^{\prime }$. The resulting equation will be in the standard form $y^{\prime }+p(x)y=q(x)$.
4. Identify $p(x)$ and $q(x)$:
   • $p(x)$ is the term multiplying $y$. Be sure to include any negative signs.
   • $q(x)$ is the term on the right-hand side of the equation.

Let's try one more example. Consider the equation:

1. Arrange: Move the $2xy$ term to the left side. $$x^2{y}^{\prime }-2xy=-5$$
2. Normalize: Divide the entire equation by $x^2$ to make the coefficient of $y^{\prime }$ equal to 1. $$\frac{x^2{y}^{\prime }}{x^2}-\frac{2xy}{x^2}=-\frac{5}{x^2}$$ Simplifying gives us the standard form: $$y^{\prime }-\frac{2}{x}y=-\frac{5}{x^2}$$
3. Identify:
   • $p(x)=-\frac{2}{x}$
   • $q(x)=-\frac{5}{x^2}$

Conclusion

In this lesson, we established the groundwork for an entire class of differential equations. You are now equipped to tackle the first and most important step in solving them.

Here are the key takeaways:

• A first-order linear differential equation can be written as $a(x)y^{\prime }+b(x)y=c(x)$. Linearity applies to the dependent variable $y$ and its derivative $y^{\prime }$, not necessarily the functions of $x$.
• The standard form is $y^{\prime }+p(x)y=q(x)$, where the coefficient of $y^{\prime }$ is 1.
• To convert an equation to standard form, you must perform algebraic manipulations—primarily division—to isolate $y^{\prime }$ with a coefficient of 1.
• Identifying $p(x)$ and $q(x)$ correctly is essential for the next steps in the solution process.

Preview of the next lesson:
In our next session, we will answer the question: "Why is identifying $p(x)$ so important?" We will introduce a concept called the integrating factor, which is derived directly from $p(x)$ and is the key to unlocking the solution to these equations.