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.

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 $y=e^{-x}$.

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 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 $g(k)=0$.
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 $p$.