First-order differential equation

Definition

Formal description

The term first-order differential equation is used for any differential equation whose order is 1. In other words, it is a differential equation of the form:

$F(x,y,y^{\prime })=0$

where $F$ is an expression (function) involving three variables. Note that $F$ must make use of $y^{\prime }$ (also written as $dy/dx$), but it could ignore $x$ or $y$.

The theory and terminology follows that for the general concept of differential equation.

Solution concept

• Functional solution: A function $f$ on the domain of interest is said to be a solution (or functional solution) to the equation if, when we plug in $y=f(x)$, the equation holds true for all $x$ in the domain, i.e.:

$F(x,f(x),f^{\prime }(x))=0\mathrm{\forall }x\in \mathrm{d}\mathrm{o}\mathrm{m}(f)$

Note that in cases of functions defined on closed intervals, we exclude checking the conditions on the boundary of the domain because two-sided derivatives don't make sense at the boundary.

• Relational solution: A relation $R(x,y)=0$ is termed a relational solution to the equation if $F(x,y,y^{\prime })=0$ holds true for all $x,y$ if we calculate the derivative $y^{\prime }$ using implicit differentiation.

Initial value problem

An initial value problem is a first-order differential equation

$F(x,y,y^{\prime })=0$

along with a point $(x_0,y_0)$.

A functional solution to the initial value problem is a solution $y=f(x)$ to the first-order differential equation such that $f(x_0)=y_0$.

A relational solution to the initial value problem is a solution $R(x,y)=0$ to the first-order differential equation such that $R(x_0,y_0)=0$.

Terminology

Solution terminology

Term	Meaning	Example
particular solution	a function or relation that is a solution for the equation (see #Solution concept). A solution in the form of a function $y=f(x)$ is termed a functional solution and a solution in the form of a relation $R(x,y)=0$ is termed a relational solution.	$y=\sin x$ is a functional solution to $y^2+y{\text{'}}^2=1$.
solution family	a family of functions or relations, with one or more parameters possibly subject to some constraints, such that for every choice of parameter values subject to those constraints, we get a particular solution.	$y=\sin (x+C)$ with parameter $C\in \mathbb{R}$, is a solution family for $y^2+y{\text{'}}^2=1$.
general solution	a solution family that covers all solutions (or almost all solutions, possibly excluding some exceptions)	The general solution to $y^{\prime }=0$ is $y=C,C\in \mathbb{R}$.
solution to initial value problem	a particular solution that satisfies the initial value condition.	A particular solution to $y+y^{\prime }=0$ satisfying $y(0)=1$ is $y=e^{-x}$.

Facts

• As a general principle, the way to solve a first-order differential equation is to convert it to an integration problem. The additive $+C$ appearing in the indefinite integration gives the freely varying parameter for the solution family.
• As a general principle, the number of degrees of freedom (i.e., the number of independent freely varying parameters) in the general solution to a first-order differential equation is 1. In other words, we expect the solution space to be one-dimensional.
• As a general principle, the number of solutions to an initial value problem should be finite. If the differential equation is nice enough, then there should be a unique solution to any initial value problem.

Geometric description of solutions

This geometric description is qualitatively different for first-order differential equations compared to higher-order differential equations.

A solution curve to a differential equation is a curve in the $xy$-plane corresponding to any solution to the differential equation. For instance, if $R(x,y)=0$ is a relational solution, then the curve $R(x,y)=0$ gives a solution curve for the differential equation.

For first-order differential equations, we generally expect that specifying a point on the curve uniquely determines the solution curve (because that's an initial value specification). Thus, the solution curves are expected to be (mostly) non-intersecting curves that (hopefully) cover the plane or a large part of the plane. This general expectation is likely to be met for first-order first-degree differential equations. For first-order differential equations of degree $n$, we generally expect up to $n$ solution curves through a generic point.

Solution strategies

Solution strategies in particular cases

Below are some formats of equations for which general strategies are known. Note that the letter $f$ is no longer used for the solution function but may be used for other functions.:

Equation type	Degree (if polynomial in highest order derivative)	Quick summary of solution strategy
first-order linear differential equation which in simplified form looks like $y^{\prime }+p(x)y=q(x)$	1	Use the integrating factor $e^{H(x)}$ where $H^{\prime }=p$. The general solution is $y=C{e}^{-H(x)}+e^{-H(x)}\int p(x)e^{H(x)}dx$
separable differential equation which is of the form $y^{\prime }=f(x)g(y)$ (any first-order first-degree autonomous differential equation is separable, though there are separable differential equations that aren't autonomous)	1	Separate and solve as $\int \frac{dy}{g(y)}=\int f(x)dx$. Also find solutions corresponding to $y=k$ where $g(k)=0$.
first-order exact differential equation $F(x,y,y^{\prime })=0$	1	Try to find a relation $R(x,y)$ such that $F(x,y,y^{\prime })=\frac{d}{dx}[R(x,y)]$ using implicit differentiation. Finding the $R$, even if it does exist, can be tricky.
Bernoulli differential equation $y^{\prime }+p(x)y=q(x)y^n$ ($n\ne 0,1$)	1	Divide both sides by $y^n$ (set aside possible stationary solution $y=0$), then substitute $w=1/y^{n-1}$ to get a first-order linear differential equation with dependent variable $w$ and independent variable $x$.
Clairaut's equation which is of the form $y=x{y}^{\prime }+f(y^{\prime })$	need not be polynomial; if polynomial, may have any degree	$y=Cx+f(C)$ with $C\in \mathbb{R}$ (all straight lines) and a single other solution explicitly described as the solution to $x+f(dy/dx)=0$, given by $x=-f^{\prime }(p),y=f(p)-p{f}^{\prime }(p)$ as a parametric curve in terms of $p$.
Lagrange equation $y=f(y^{\prime })x+g(y^{\prime })$ which is linear in $x$ and $y$ but not necessarily in $y^{\prime }$	need not be polynomial; if polynomial, may have any degree	General solution is a family of curves, each described as a parametric curve with parameter the derivative $y^{\prime }$ (which we denote by $p$). There may be some special straight line solutions of the form $y=px+g(p)$ for values $p$ with $p=f(p)$.

Reduction methods

The following are methods that may be used to reduce more complicated first-order differential equations to simpler ones:

Method	Can it reduce the degree of the differential equation, and potentially convert a higher degree differential equation to first-degree differential equations?	Description
substitution method for solving differential equations (the substitutions of interest are zeroth-order substitutions)	No	We put $u=g(x,y)$ and replace one of the variables $x,y$ with $u$. Then, after solving, we plug back $u=g(x,y)$ to get the relational solutions between $x$ and $y$.
factorization method for solving differential equations	Yes	Bring everything to one side, factor it as a product of two expressions, then solve each as a separate differential equation, and combine solutions. Also, check for mixed solutions.