Table of Contents
Understanding First-Order Differential Equations
In this chapter we focus only on first-order differential equations: equations where the highest derivative that appears is the first derivative.
A general first-order differential equation involving an unknown function $y$ of $x$ can be written as
$$
\frac{dy}{dx} = f(x, y),
$$
or equivalently
$$
y' = f(x, y).
$$
The “order” is $1$ because only $y'$ (the first derivative) appears, and no higher derivatives like $y''$.
A solution is a function $y(x)$ that, when you compute its derivative $y'(x)$ and substitute both $y(x)$ and $y'(x)$ into the equation, makes the equation true for all $x$ in some interval.
Much of the work with first-order equations consists of:
- Recognizing a special type of first-order equation, and
- Applying a matching method to solve it.
This chapter gives a basic overview of several common special types and their solution ideas, except for separable equations, which will be treated in its own section later.
General vs. special forms
The most general first-order equation,
$$
y' = f(x, y),
$$
is typically too broad to “solve in general.” Instead, we usually look for recognizable forms such as:
- Linear first-order equations
- Separable equations
- Exact equations
- Homogeneous first-order equations (in a specific sense)
- Simple equations of the form $y' = f(x)$ or $y' = f(y)$
Each type has its own solving technique. Some equations fit into more than one category; that simply gives you more than one way to solve them.
In an introductory setting, the primary goals are:
- Being able to identify the type of first-order equation from its form, and
- Knowing what general solving strategy goes with each type.
Linear first-order differential equations
A very important and common type is the linear first-order equation in $y$:
$$
y' + p(x)\,y = q(x),
$$
where $p(x)$ and $q(x)$ are known functions of $x$ (they do not depend on $y$).
“Linear in $y$” means that:
- $y$ and $y'$ appear only to the first power,
- there are no products like $y^2$, $(y')^2$, $y \cdot y'$, or $\sin(y)$, etc.
For instance:
- $y' + 3y = 6x$ is linear,
- $y' - \dfrac{2}{x}y = x^2$ is linear,
- $y' + y^2 = x$ is not linear (because of $y^2$).
Linear first-order equations are especially important because:
- They model many real-world processes (cooling, simple population models with external input, RC circuits, etc.).
- They always have a systematic method of solution based on rewriting the equation in a certain way (often via an integrating factor, treated in more detail elsewhere).
Recognizing linear vs. nonlinear
To decide if
$$
y' = f(x, y)
$$
is linear in $y$, try to rewrite it in the form
$$
y' + p(x)\,y = q(x).
$$
If this is possible with $p$ and $q$ depending only on $x$, then it is a (first-order) linear equation.
If $y$ appears in a nonlinear way—like $y^2$, $e^y$, $\sin y$, or multiplied with $y'$—then the equation is nonlinear.
For example:
- $y' = x^2 - 4y$ can be rearranged as $y' + 4y = x^2$ → linear.
- $y' = x y^2$ cannot be rearranged to $y' + p(x)\,y = q(x)$ → nonlinear.
Separable first-order equations (overview only)
A very useful special type, studied in its own chapter, is the separable equation. It is an equation that can be written in the form
$$
\frac{dy}{dx} = g(x)\,h(y),
$$
so that you can “separate variables”:
$$
\frac{1}{h(y)}\,dy = g(x)\,dx,
$$
and then integrate both sides with respect to their own variables.
You will learn the details of this method, along with step-by-step examples, in the dedicated section on separation of variables. Here you only need to recognize the form: the right-hand side factors into a function of $x$ times a function of $y$.
Examples:
- $y' = x y$ is separable: $g(x) = x$, $h(y) = y$.
- $y' = \dfrac{y}{x}$ is separable: $g(x) = \dfrac{1}{x}$, $h(y) = y$.
- $y' = x + y$ is not separable because $x + y$ does not factor into $g(x)h(y)$.
Equations of the form $y' = f(x)$
A very simple subclass is when the derivative depends only on $x$:
$$
\frac{dy}{dx} = f(x).
$$
In this case, the general solution is obtained just by antidifferentiating:
$$
y(x) = \int f(x)\,dx + C,
$$
where $C$ is a constant of integration.
This is the differential-equation way of looking at basic antiderivatives: solving $y'(x) = f(x)$.
For example:
- If $y' = 3x^2$, then $y(x) = x^3 + C$.
- If $y' = e^{2x}$, then $y(x) = \dfrac{1}{2}e^{2x} + C$.
This is a special case of both:
- a separable equation, and
- a linear first-order equation with $p(x)=0$.
Equations of the form $y' = f(y)$
Another simple pattern is
$$
\frac{dy}{dx} = f(y),
$$
where the derivative depends only on $y$.
These equations are automatically separable because you can rewrite them as
$$
\frac{1}{f(y)}\,dy = dx,
$$
and then integrate both sides. Precisely how to carry this out depends on $f(y)$ and is treated under separation of variables.
From the point of view of recognizing equations, remember:
- If the right-hand side is only a function of $y$, you have $y' = f(y)$, which is separable and therefore solvable (at least in principle) by separation of variables.
Exact first-order equations (conceptual overview)
Some first-order equations can be written in the form
$$
M(x, y)\,dx + N(x, y)\,dy = 0.
$$
If such an equation comes from the derivative of some underlying function $F(x, y)$, we say the equation is exact. In that case, there is a function $F$ with
$$
\frac{\partial F}{\partial x} = M(x, y), \quad
\frac{\partial F}{\partial y} = N(x, y),
$$
and the solutions are given implicitly by
$$
F(x, y) = C.
$$
Recognizing whether an equation is exact, and how to find $F(x, y)$, requires tools from multivariable calculus (partial derivatives). At an introductory level, the main idea to remember is:
- An exact first-order equation is one where the differential expression comes from a “potential function” $F(x, y)$.
- Solving it reduces to finding $F$ and then setting $F(x, y)=C$.
The detailed condition for exactness and the practical solving steps are usually treated in a more advanced or separate chapter.
Homogeneous first-order equations (in a special sense)
There is another use of the word “homogeneous” for first-order equations (different from “homogeneous linear equation” in higher-order contexts).
A first-order equation
$$
\frac{dy}{dx} = f\!\left(\frac{y}{x}\right)
$$
is called homogeneous (in this special sense) if the right-hand side depends only on the ratio $y/x$.
Such equations can be transformed by the substitution
$$
v = \frac{y}{x}, \quad \text{so } y = vx,
$$
and then
$$
y' = v + x\,v',
$$
which turns the original equation into one involving $v$ and $x$ only. Often this transformed equation becomes separable in $v$ and $x$.
The key points at this stage:
- Recognize that an equation is homogeneous of degree zero if its right-hand side is a function of $y/x$ alone.
- Know that the usual solving idea is to substitute $y = vx$ (or $y = ux$, etc.) to reduce it to a simpler first-order equation in the new variable.
Example of the form:
$$
\frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x} = 1 + f\!\left(\frac{y}{x}\right),
$$
which can be rearranged into a form suitable for the substitution $y = vx$.
Initial value problems (IVPs) for first-order equations
A first-order initial value problem (IVP) is a first-order equation together with a condition that specifies the value of the solution at one point:
$$
\frac{dy}{dx} = f(x, y), \quad y(x_0) = y_0.
$$
Here:
- The differential equation gives a family of solutions containing an arbitrary constant $C$.
- The initial condition $y(x_0) = y_0$ is used to determine the particular value of $C$ and thus pick out one specific solution curve.
For example, if $y' = 3x^2$ has the general solution $y = x^3 + C$, then the IVP
$$
y' = 3x^2, \quad y(1) = 5
$$
leads to
$$
5 = 1^3 + C \quad \Rightarrow \quad C = 4,
$$
so the unique solution to the IVP is $y = x^3 + 4$.
The existence and uniqueness of solutions for first-order IVPs (under certain conditions on $f$) is a theoretical topic usually discussed in more advanced sections, but the practical procedure is straightforward:
- Solve the differential equation in general form,
- Plug in the initial condition,
- Solve for the constant.
Geometric view: direction fields
Even without explicit formulas for solutions, first-order equations can be visualized using a direction field (also called a slope field).
Given
$$
y' = f(x, y),
$$
at each point $(x, y)$ in the plane, you can:
- Compute the slope $m = f(x, y)$,
- Sketch a short line segment through $(x, y)$ with slope $m$.
Repeating this at many points gives a direction field. A solution curve $y(x)$ then appears as a curve that “follows” the small line segments everywhere—its tangent at each point matches the local direction indicated by the field.
This geometric picture helps you:
- Understand the qualitative behavior of solutions (e.g., do they increase or decrease? Do they level off?),
- Compare different solutions with different initial conditions,
- See features such as equilibrium or steady-state solutions, which occur when $y' = 0$.
Direction fields are especially useful when a formula for the solution is difficult or impossible to write down using elementary functions.
Summary of key patterns
For first-order differential equations, the central skill is recognizing forms:
- $y' = f(x)$
→ Integrate $f(x)$ with respect to $x$. - $y' = f(y)$
→ Separable: rewrite $\dfrac{dy}{f(y)} = dx$. - $y' = g(x)\,h(y)$
→ Separable: rewrite $\dfrac{1}{h(y)}\,dy = g(x)\,dx$. - $y' + p(x)\,y = q(x)$
→ Linear first-order; has a systematic method (using an integrating factor). - $M(x, y)\,dx + N(x, y)\,dy = 0$ and exact
→ Comes from a function $F(x, y)$ with $F(x, y) = C$ as the solution. - $y' = f\!\left(\dfrac{y}{x}\right)$
→ Homogeneous (in the special sense); use substitution $y = vx$.
Each of these special types will be explored in more detail, with concrete solving techniques and worked examples, in the dedicated subsections that follow, especially in the chapter on separation of variables.