# Preface

This section is devoted to a review of the basic terminology and definitions of solutions to differential equations. Since this topic becomes a substantial part of functional analysis and operator theory, this section also addresses this point of view in layman's terms.

# Solutions to ODEs

The first order differential equation $$R(x,y,y' ) = 0$$ is an equation containing a derivative of an unknown function. It can be also written in the equivalent form involving differentials $$G \left( x,y,{\text d}x. {\text d}y \right) = 0 .$$ The derivative of any function is the result of applying the derivative operator to the function. This can usually be denoted by using Euler's notation $$\texttt{D} = {\text d}/{\text d}x ,$$ which is an unbounded (not continuous) operator because it may map a bounded function into unbounded one. The operator D does not act on arbitrary functions, such as the absolute value function. The function must be smooth so that the derivative operator can be applied. The derivative operator is defined on some subset of continuous functions. So, the domain of the derivative operator includes only continuously differential functions on some open interval (𝑎,b) that we denote as C¹(𝑎,b) or simply C¹. Because D is unbounded, every differential equation should be considered on a set of functions that have a continuous derivative on some open interval.

In calculus, you learn that the simplest differential equation Dy = f(x) has infinitely many solutions $$y(x) = \int f(x)\,{\text d}x +c ,$$ depending on an arbitrary constant c, because the derivative operator D annihilates constants. So the null-space of the derivative operator is a one-dimensional space spanned on any nonzero constant function. (Remember: the derivative of any constant, regardless of its value, will always be zero.) For example, the differential equation y' = x-1 has the solution $$y(x) = \int x^{-1}\,{\text d}x + c = \ln |x| + c,$$ where c is any real constant. However, we can represent this constant as logarithm c = lnC, where C is some positive number. Then our solution becomes $$y(x) = \ln (c\,|x|) = \ln C\,x ,$$ where C is a positive constant when x is positive and C becomes negative for negative independent variable x. Therefore, in our formula, C is an arbitrary constant, but taking from some domain depending on the independent variable. This example shows that an arbitrary constant that is used to represent a solution of a differential equation should be chosen with care from some domain depending on the formula in use.

Now we discuss a topic of inverse differential operators that we started previously in the section. This means that the inverse operator D-1 does not exist and the domain of D must be more restricted in order to find its inverse. Therefore, you expect that a first order differential equation also has infinitely many solutions depending on an arbitrary constant. It turns out that this is usually the case (with some exceptions), but an arbitrary constant C is not 100% arbitrary and should be taken from some domain depending on the problem and its possible solution. It is a custom to call such family of solutions depending on a constant, the general solution, leaving the word antiderivative to calculus.

A general solution of a first-order differential equation $$F(x,y,y' ) = 0$$ is a family of solutions containing an arbitrary constant (from some domain). A particular solution is derived from the general solution by setting the constant to particular value, often chosen to fulfill an initial condition.
As a matter of terminology, we mention that in different applications solutions of differential equations are also called integrals or streamlines. The set of points satisfying an equation $$F(x,y,c ) = 0$$ is called the locus. The word "locus" is Latin for "place" and refers to the places in the (x,y)-plane where the equation is true. For example, the locus defined by x² + y² = 4 is a circle of radius 2 centered at the origin.
 Some solution curves for the differential equation y' = y - x². > > > DSolve[{y'[x] == y[x] - x^2, y[0] == a}, y[x], x] f[a_, x_] = 2 - 2 E^x + a E^x + 2 x + x^2; parameters = {0, 0.1, 0.2, 0.5, 1}; family = Plot[Evaluate[f[#, x] & /@ parameters], {x, -2, 3.3}, PlotRange -> {-2, 3}, Axes -> False]; a1 = Graphics[{Black, Arrowheads[0.06], Arrow[{{-2, 0}, {3.3, 0}}]}]; a2 = Graphics[{Black, Arrowheads[0.06], Arrow[{{0, -2}, {0, 3}}]}]; ll = Graphics[{Thick, Line[{{1.85, 3}, {2.8, 1.0}}]}]; p = Graphics[{PointSize[Large], Purple, Point[{2.13, 2.38}]}]; t1 = Graphics[ Text[StyleForm["(" Subscript[x, 0], FontSize -> 14], {2.4, 2.5}]]; t1a = Graphics[Text[StyleForm[",", FontSize -> 16], {2.66, 2.5}]]; t1b = Graphics[ Text[StyleForm[Subscript[y, 0], FontSize -> 14], {2.84, 2.5}]]; t1c = Graphics[Text[StyleForm[")", FontSize -> 14], {3.05, 2.5}]]; t2 = Graphics[ Text[StyleForm["y'(" Subscript[x, 0], FontSize -> 14], {3.1, 1.1}]]; t2a = Graphics[Text[StyleForm[")", FontSize -> 14], {3.39, 1.1}]]; tx = Graphics[ Text[StyleForm["x", FontSize -> 14, FontWeight -> "Bold"], {3.2, 0.35}]]; ty = Graphics[ Text[StyleForm["y", FontSize -> 14, FontWeight -> "Bold"], {0.25, 2.75}]]; Show[family, a1, a2, ll, p, t1, t1a, t1b, t1c, t2, t2a, tx, ty]

The situation is similar to some practical problems from physics and mechanics. Suppose we have a single railcar (engine) moving along a track that we can describe by the function x(t). If we express its velocity as a function of time and position, we get the first order differential equation $$v = \dot{x} = f(t,x) ,$$ where x is a position of the engine at time t. Its position x(t) has one degree of freedom because the location of the car is defined by the distance along the track.

A solution is a relation between the variables (independent and dependent), which is free of derivatives of any order, and which satisfies the differential equation identically in some interval.
Most differential equations do not have solutions that can be written in elementary form, and even when they do, the search for formulas often obscures the central question: How do solutions behave? Generally speaking, there is no better way to define a solution than to say that it satisfies the differential equation. Nevertheless, there are two specific representations that are very important.
If a differential equation $$F(x,y,y' ) =0$$ has a solution that is represented via a known smooth function y = φ(x) on some interval (𝑎,b), then this function φ is known as an explicit solution. A relation ψ(x,y) = 0 is known as the implicit solution of the given differential equation if it defines at least one real function of the variable x on an interval (𝑎,b) such that this function is an explicit solution of the differential equation on this interval.
Recall that we can derive a differential equation from the relation c = ψ(x, y), or more general Ψ(x, y, c) = 0, involving an arbitrary constant c, by simply differentiating the relation to eliminate the constant c. If a differential equation can be derived with such eliminating procedure, then the relation Ψ(x, y, c) = 0 defines the solution in implicit form. Such relation is possible mostly in theoretical situations. A transition from the implicit solution to the explicit solution provides the following theorem.
The Implicit Function Theorem: Consider a continuously differentiable function F: ℝ² → ℝ and a point (x0,y0) ∈ ℝ² so that F(x0,y0) = c, a constant. If $$\frac{\partial F}{\partial y} \left( x_0 , y_0 \right) \ne 0,$$ then there is a neighborhood of (x0,y0) so that whenever x is sufficiently close to x0, there is a unique y so that F(x,y) = c. Moreover, this assignment makes y continuous function of x.
Example 1: While you are not yet familiar with methods needed to solve the ordinary differential equations, there are some very simple first and second order equations that can be solved by inspection. One of these is
$y' = 2\,y .$
This first order differential equation asks you to find a function y = ϕ(x) that is equal to half of its own derivative at every value of x. From calculus you know that the derivative of the exponential function is $$\frac{{\text d}}{{\text d}x} \, e^{kx} = k\, e^{kx} .$$ So we immediately determine that k = 2 and we get the solution $$y = \phi (x) = e^{2\,x} .$$ Further inspection reveals that there is a family of solutions $$y = C\, e^{2\,x} ,$$ depending on an arbitrary real constant C ∈ ℝ. However, we can represent this constant as $$C = e^c ,$$ some real constant c. This leads to another family of solutions $$y = e^{x+c} .$$ It turns out that arbitrary constant c can be not only real, but also a complex number.

Suppose we want to find a solution that satisfies the initial condition y(0) = -1. This leads to the equation $$e^c = -1 ,$$ that tells us that $$c = {\bf j}\pi + 2k\,{\bf j}\pi ,$$ for k = 0, ±1, ±2, … , where j is the unit vector in the positive direction on the complex plane ℂ. Although the given initial value problem has a unique solution, it has infinitely many ways to write it down. Therefore, representing the family of solutions in the form $$y = e^{x+c}$$ yields complex choice for an arbitrary constant c.

Our next differential equation

$y'' = 4\,y$
is of the second order. Now we have two solutions: $$y = x^{2x}$$ and $$y = x^{-2x} .$$ from which we build a two-parameter family of solutions
$y = C_1 e^{2x} + C_2 e^{-2x} .$
It is natural for the second order differential equation to have a two-parameter family of solutions. However, we can use other functions, called hyperbolic trigonometric functions
$\cosh (kx) = \frac{1}{2} \left( e^{kx} + e^{-kx} \right) \qquad\mbox{and} \qquad \sinh (kx) = \frac{1}{2} \left( e^{kx} - e^{-kx} \right) .$
Then the given second order differential equation admits another family of solutions:
$y = \phi (x) = c_1 \cosh (2x) + c_2 \sinh (2x) ,$
where c1, c2 are arbitrary real constants.    ■

Since the general solution of the differential equation is usually not possible to determine, we need a way to pin down a particular solution. The most obvious possibility to identify a particular solution is to impose the initial condition by specifying the value of a solution at some point. The corresponding problem, which is called the initial value problem, consists of some differential equation together with an initial condition:

$F(x,y,y' ) = 0 , \qquad y(x_0 ) = y_0 ,$
where x0 and y0 are some specified real numbers.

Members of the family of solutions that form the general solution do not intersect and do not touch each other. Why? If two members of the general solution will have a common point, then at this point we would have two solutions of the same equation written in implicit form. However, there could be other solutions that may touch some members of the general solution. These exceptional solutions are called singular solutions.

General Solution

Let us consider a first order differential equation written in normal form (so the derivative is isolated):
$$\label{EqSolution.1} y' = f(x,y) \qquad \mbox{or} \qquad \frac{{\text d}y}{{\text d}x} = f(x,y) \qquad \mbox{or} \qquad \texttt{D}y = f(x,y) ,$$
where $$\texttt{D} = {\text d}/{\text d}x$$ is the derivative operator (unbounded). A particular case of this equation you studied in calculus when f = f(x). independent of y. You learn that this differential equation has infinite many solutions depending on an arbitrary constant C (from some domain). All these solutions are referred to as antiderivative. In other word, $$\texttt{D}^{-1} f(x) = \int f(x) \,{\text d}x -C$$ includes the one-dimensional null-space $$\texttt{D}^{-1} f(x,y) ,$$ depending on a parameter C.

It is expected that we observe a similar property in general case of Eq.\eqref{EqSolution.1}. Unfortunately, in this case the null-space of the inverse operator $$\texttt{D}^{-1}$$ applied to function f(x, y) may be more complicated than just one dimensional space. Therefore, in the theory of differential equation the word antiderivative is not used because it does not reflect the general situation. Instead, we use expression general solution that means a one-dimensional subspace of the null-space $$\texttt{D}^{-1} f(x,y) .$$

Let us look at formation of the differential equation from another prospective. Suppose we are given a one-dimensional family of smooth functions F(x, y, c) = 0, where c is a real parameter. Upon differentiating this relation we get

$F(x,y,c) = 0 \qquad\mbox{and} \qquad \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}\, y' = 0.$
Assuming that ∂F/∂y ≠ 0, we exclude parameter c from these two equations, and obtain a differential equation R(x, y, y') = 0, for which the family of solutions F(x, y, c) = 0 provide a one-dimensional family of solutions. This mafily of integral curves for the differential equation F(x, y, c) = 0 is referred to as the primitive of the differential equation.
Example 2: Consider a family of conic curves
$\frac{x^2}{a^2 + c} + \frac{y^2}{b^2 + c} = 1 ,$
where 𝑎 and b are definite constants and is an arbitrary parameter. This equation describes a family of confocal conics.
 parameters = {0, 0.5, 1.0, 2.05, 5.1};; ContourPlot[ Evaluate[f[x, y, #] == 1 & /@ parameters], {x, -2.5, 2.5}, {y, -4, ; 4}, AspectRatio -> 1, PlotRange -> {-2, 2}, ; PlotLegends -> Table[Row[{"c=", j}], {j, parameters}]] Confocal conics. Mathematica code

The differential equation that is primitive is obtained by eliminating c between it and the derivative equation
$\frac{2x}{a^2 +c} + \frac{2y\,y'}{b^2 +c} = 0.$
From the primitive and the derived equation it is found that
$a^2 + c = \frac{x^2 y' - xy}{y'} , \qquad b^2 + c = y^2 -xy\,y' ,$
and eliminating parameter c
$a^2 -b^2 = \frac{x^2 y' - xy}{y'} - y^2 + xy\, y' .$
Therefore, the required differential equation becomes
$xy\,y' + \left( x^2 - y^2 - a^2 + b^2 \right) y' = xy .$
This differential equation has the general solution defined by the family of confocal conics.    ■
This procedure of differential equation formation from a primitive can be easily extended for multi-dimensional case. For example, suppose we are given a family of two parameter curves
$F(x,y,c_1 , c_2 ) = 0 .$
Upon differentiation, we obtain the system of algebraic equations
$F(x,y,c_1 , c_2 ) = 0 , \qquad \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}\, y' =0, \qquad \frac{\partial^2 F}{\partial x^2} + \frac{\partial F}{\partial y}\, y'' + \frac{\partial^2 F}{\partial y^2}\, \left( y'\right)^2 =0,$
from which two parameters c1, c2 are eliminated. This yields a second order differential equation for which the F(x, y, c1, c2) = 0 provides a family of solutions.

We can conclude that a family of curves depending on n parameters generates a differential equation for which this family provides a general solution. However, it is not evident that any differential equation of order n can be derived from such primitive. It does not follow that if the differential equation is given, it possesses a general solution that depends upon n arbitrary constants. This conclusion follows from the existence and uniqueness theorem that guarantees dependence of the general solution on n initial conditions. In the formation of a differential equation from a given family of integral curves, corresponding conditions must be assumed to be satisfied as it is required by the existence theorem. The role of n arbitrary constants can play n initial conditions---they are more natural than finding a corresponding primitive (= implicit general solution). You will learn shortly that such general solution is impossible to construct explicitly for an arbitrary differential equation even if it is known that the corresponding primitive exists.

From a theoretical point of view, a differential equation can be considered as an unbounded operator (because it contains a derivative operator) mapping a smooth function into another continuous function. Its null-space is at least n-dimensional, which is reflected by the general solution containing exactly n arbitrary constants. Therefore, a general solution of any differential equation of order n should contain exactly n arbitrary constants taken from some domain (assuming that the equation has a solution). Does it mean that any solution to the differential equation should be a member of the general solution? The answer is negative because some nonlinear differential equations may have solutions not included in the general solution. We will discuss these exceptional solutions, called the singular solutions in the next section.

Solution terminology

Term Meaning Example
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. $$y = \cosh (x + C)$$ with parameter C ∈ ℝ, is a solution family for $$y^2 - (y' )^2 =1 .$$
general solution a family of solutions including a one-dimensional subspace of the null-space depending on one arbitrary constant; it covers almost all solutions excluding some exceptions The general solution to $$y' = 1$$ is $$y = x + C ,$$ where C ∈ ℝ.
explicit solution a solution of the differential equation that is expressed as some known function y = φ(x). $$y = 2 \left( C - x^2 \right)^{-1}$$ is an explicit (general) solution to the differential equation $$y' = x\,y^2 .$$
implicit solution a solution of the differential equation in the form f(x,y) = 0 for some function f that does not depend on the derivative. $$2 = y \left( C - x^2 \right)$$ is an implicit (general) solution to the differential equation $$y' = x\,y^2 .$$
particular solution a function or relation that defines a solution for the differential equation. Usually,a particular solution is obtained from the general solution by choosing a particular value for the parameter of from the initial value problem. y = cosh x is a functional solution to $$y^2 - (y' )^2 =1 .$$
singular solution a solution of the differential equation that cannot be obtained from the general solution for any values of constant C, including infinity, and for which the initial value problem has multiple solutions. y ≡ 0 is a singular solution to the initial value problem $$y' = \sqrt{y}, \ y(0) =0 .$$
solution to initial value problem a particular solution that satisfies the initial value condition. A particular solution to $$y' + y = 1, \ y(0) =2$$ is $$y = 1 + e^{-x} .$$
Example 3: The differential equation $$y' = y^{1/2}$$ has the solution $$y = (1/4) \left( x^2 + 2x + 1 \right) = (1/4) \left( x + 1 \right)^2$$ since $$y' = (1/2) \left( x + 1 \right) .$$ This solution is defined for all real numbers. More generally, we can check that $$y' = (1/2) \left( x^2 + 2xC + C^2 \right)$$ is a solution for any C ∈ ℝ. For the initial value problem
$(y' )^2 = y , \qquad y(0 ) = 4 ,$
we can find the corresponding value of C to be $$C = \pm 4 .$$

■

Constant functions are usually not solutions of the differential equations unless they annihilate the slope function. In this case, we call them critical points or equilibrium/stationary solutions. Therefore, if y = y* vanishes the slope function, $$f(x, y^{\ast} ) \equiv 0 ,$$ then y = y* is the equilibrium solution of the differential equation $$y' = f(x, y ) .$$ A stationary solution may be singular or not, which we will show in the following examples in the next section.

1. Bougoffa, L., On the exact solutions for initial value problems of second-order differential equations, Applied Mathematics Letters, 2009, Vol. 22, pp. 1248--1251.
2. Hasan, Y.Q., A new development to the Adomian decomposition for solving singular IVP of Lane--Emden type, United States of America Research Journal (USARJ), 2014, Vol. 2, No.3, pp. 9--13.