Differential equation

Definition

Formal description

The term differential equation, sometimes called ordinary differential equation to distinguish it from partial differential equations and other variants, is an equation involving two variables, an independent variable $x$ and a dependent variable $y$, as well as the derivatives (first and possibly higher) of $y$ with respect to $x$. Formally, it is an equation of the form:

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

where $F$ is a function of $k+2$ variables. Here $k\ge 1$. Note that $F$ may choose not to use some of the derivatives.

In functional notation, the same differential equation may be written as:

$F(x,f(x),f^{\prime }(x),f^{″}(x),\dots ,f^{(k)}(x))=0$

where $f$ is the function such that $y=f(x)$.

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.:

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

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 $F(x,y,y^{\prime },y^{″},\dots ,y^{(k)})=0$ holds true for all $x,y$ if we calculate the derivatives of $y$ with respect to $x$ using implicit differentiation.

Initial value problem

An initial value problem is a differential equation:

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

accompanied with a tuple $(x_0,y_0,y_1,\dots ,y_{k-1})$.

A functional solution to the initial value problem is a functional solution $y=f(x)$ for the differential equation such that $f(x_0)=y_0$ and $f^{(i)}(x_0)=y_i$ for $i\in \{1,2,\dots ,k-1\}$.

Analogously, we can define a relational solution to the initial value problem.

Key observations

Differential equations are functional equations

Differential equations are examples of functional equations. A functional equation is an equation where the variable that we are trying to solve for is a function, and the equation holds true for all values of the input to the function. For instance, here is an example of a functional equation (that's not a differential equation):

$f(x+y)=f(x)+f(y)\mathrm{\forall }x,y\in \mathbb{R}$

A solution to a functional equation is a function that satisfies the equation for all choices of inputs. For instance, any function of the form $f(x):=ax$ for fixed $a\in \mathbb{R}$ is a solution to the above functional equation.

Differential equations are functional equations -- we are trying to solve a differential equation, not for the variables, but for the functional relationship between them.

Differential equations capture behavior at a single point

Not every functional equation involving derivatives is a differential equation. Differential equations are characterized by the evaluation of the function and its derivatives all happening at a single point. For instance:

• $x^2=f(x)+(f^{\prime }(x){)}^3$ is a differential equation because all the function and derivative evaluations happen at $x$, but
• $x^2=f(x)+f^{\prime }(1-x)$ is not a differential equation in our sense of the word because the derivative evaluation happens at $1-x$ rather than $x$.

Another way of putting this is that differential equations are inherently local and cannot relate the behavior of the function at far-away points.

Functional equations involving derivatives that do not fit this definition of differential equation are also studied, but the study of these is more complicated and requires new techniques. Delay differential equations is one such class of functional equations.

It does not make sense to ask whether a point satisfies a differential equation

Consider a differential equation:

$x^2=y+y^{\prime }$

If I ask the question: does the point $x=2,y=3$ satisfy the differential equation?, the answer is that the question doesn't make any sense. This is because verifying a differential equation requires knowing the functional relationship between $x$ and $y$, which in turn allows us to compute the numerical value of $y^{\prime }$ and check whether the equation is satisfied.

Terminology

Equation terminology

Term	Meaning	Example (don't try to solve these differential equations!)
order of a differential equation	it is the largest $k$ for which the $k^{th}$ derivative of the dependent variable appears in the differential equation.	The equation $y+x{y}^{‴}+(y^{″}{)}^2=\sin (y^{\prime })$ has order three because $y^{‴}$ is the largest derivative appearing.
first-order differential equation	differential equation of order one, i.e., it involves only $x,y,y^{\prime }$.	$y^{\prime }=\sin (x+y{y}^{\prime })$ is a first-order differential equation.
second-order differential equation	differential equation of order exactly two, i.e., it involves only $x,y,y^{\prime },y^{″}$ and has at least one appearance of $y^{″}$.	$x{y}^{″}+\cos (y^2{y}^{\prime })=x^2{e}^{y{y}^{\prime }}$ is a second-order differential equation.
degree of a differential equation	if the differential equation is polynomial in terms of its highest order derivative, then the degree of that polynomial.	$(y^{‴}{)}^2{y}^{\prime }+(y^{\prime }{)}^5=3xy$ has degree two.
autonomous differential equation	differential equation where the independent variable does not appear explicitly anywhere in the equation.	$y+y^{″}=\cos (y{y}^{\prime }{y}^{‴})$ is autonomous. On the other hand, $y+x{y}^{\prime }=y^{″}$ is not autonomous
linear differential equation	A differential equation of the form $p_k(x)y^{(k)}+p_{k-1}(x)y^{(k-1)}+\dots +p_0(x)y=q(x)$ where the $p_i$s and $q$ are all functions. In other words, the expression is linear in $y$ and its derivatives with coefficients in terms of $x$. Linear differential equations are usually written with the coefficient of $y^{(k)}$ cleared to 1, by dividing throughout by the coefficient of $y^{(k)}$.	$e^x{y}^{‴}+x^2{y}^{″}+\sin (x)y^{\prime }+3y=x^2-2x+5$ is linear.
homogeneous linear differential equation	A linear differential equation of the form $p_k(x)y^{(k)}+p_{k-1}(x)y^{(k-1)}+\dots +p_0(x)y=0$. In other words, the constant term function is zero.	$e^x{y}^{(4)}-3x^3{y}^{‴}+\sin (\sin x)y=0$ is homogeneous linear.
linear differential equation with constant coefficients	A linear differential equation of the form $a_k{y}^{(k)}+a_{k-1}{y}^{(k-1)}+\dots +a_1{y}^{\prime }+a_{0}y=b$ where all the $a_i$s and $b$ are zero.	$2y^{(5)}-y^{(3)}+y=13$ is linear with constant coefficients.

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 $y+y^{\prime }+y^{″}=(x+1{)}^2$ satisfying $y(0)=-1,y^{\prime }(0)=0$ (i.e., $y=-1$ and $y^{\prime }=0$ when $x=0$) is $y=x^2-1$.

Facts

• As a general principle, the way to solve a differential equation of order $k$ is to reduce it to a sequence of $k$ integration problems. Each integration problem introduces a new freely varying parameter.
• 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 differential equation of order $k$ must equal $k$. There are various exceptions and irregularities, but this is what we should generally expect. Another way of putting this is that the solution space to a differential equation of order $k$ is expected to be $k$-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.

Solution strategies

General idea for strategy toward a general solution

The rough idea is to convert the differential equation to a sequence of integration problems. If the differential equation has order $k$, it should reduce to performing $k$ integrations. Each integration introduces a new freely varying parameter.

CAUTION: There may be other auxiliary integrations that need to be done, e.g., for computation of integrating factors or in order to solve an integration by parts problem. These auxiliary antiderivative computations do not, however, introduce new freely varying parameters.

We see that this general strategy may run into trouble at many levels:

Stage	Type of difficulty
Differential equation to integration	There is no general-purpose algorithm for converting an arbitrary differential equation to an integration problem or sequence of integration problems. Thus, we may not even be able to get started on the process. For some types of structures of differential equations, strategies are known for converting them to integration problems. For others, we have to use ad hoc techniques. There exist differential equations for which there is no way of converting them to integration problems. Note that for differential equations of order two or higher, this problem may occur at any of the stages, i.e., we may be able to do one level of integration but not the next.
Solving the integration problem to get rid of the integral sign	Even after we've converted the differential equation to an integration problem, there may not be any analytic methods for computing the antiderivative. Note that this is not such a big issue because there are known techniques for calculating approximate solutions to integration problems even if analytical methods are not available for getting a precise solution.
Making sense of relational solutions, converting to functional solutions or at least trying to understand what they mean	Even after we have done the desired antidifferentiations, the solutions may be in relational form rather than functional form. This means that we may still not have explicit functional descriptions of $y$ in terms of $x$. We may not even know whether $y$ is expressible as a function of $x$. We may not even know if the relational solution has any points in it at all!

Going from a particular solution to a general solution

There are some special types of differential equations where, once we find a particular solution, we can find other solutions, perhaps even the general solution. Some notable cases are considered here:

• For a linear differential equation, finding a general solution is equivalent to finding a particular solution + solving the corresponding homogeneous linear differential equation. In particular, for a linear differential equation with constant coefficients, there is a closed form expression for the solution of the corresponding homogeneous linear differential equation with constant coefficients, so finding a particular solution is equivalent to finding the general solution.
• For an autonomous differential equation, if $y=f(x)$ is a solution, so is $y=f(x+C)$ for any constant $C$. Note that this does not give the general solution if the order is more than one, but it does help move from a particular solution to a solution family with one parameter.

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	Order	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	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	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 $y=x{y}^{\prime }+f(y^{\prime })$	1	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 $p$.
second-order autonomous differential equation of degree one, which is of the form $y^{″}=F(y,y^{\prime })$	2	1