Differential Equations

Introduction

1.A differential equation is an equation whose unknown is a function and which contains one or more derivatives of that function. By contrast, the unknowns in an algebraic equation are numbers. For a function $y = y(x)$ of one variable, a differential equation of order $n$ has the implicit form:

$$F\left(x, y, y', y'', \ldots, y^{(n)}\right) = 0$$

2.The dependence of $F$ on $y^{(n)}$ means that the derivative of order $n$ is present. A solution on an interval $I$ is a function $y:I \to \mathbb{R}$ for which all required derivatives exist and the equation holds at every point of $I.$ The domain is part of the solution because the equation and the required derivatives must be defined throughout $I.$ For example, $\ln|x| + C$ solves $y' = 1/x$ on an interval that does not contain $0,$ but it is not a solution on an interval that contains $0.$

3.A first-order example is:

$$y' + y = 0$$

4.For each real number $C,$ the function $y(x) = Ce^{-x}$ has derivative $y'(x) = -Ce^{-x}.$ Substitution gives:

$$y'(x) + y(x) = -Ce^{-x} + Ce^{-x} = 0$$

5.Conversely, if $y$ is any solution, the product rule gives $\left(e^xy\right)' = e^x(y' + y) = 0.$ Hence $e^xy = C,$ and $y = Ce^{-x}.$ The family $y(x) = Ce^{-x},$ with $C \in \mathbb{R},$ is the general solution on $\mathbb{R}.$ A single member of the family is a particular solution. An initial condition may select one member.

6.For a positive solution, $-y' = y,$ so the rate of decrease is proportional to the current value. The condition $y(0) = y_0 > 0$ gives the particular solution $y(x) = y_0e^{-x},$ whose value is halved over each interval of length $\ln(2).$ Models of radioactive decay and capacitor discharge have the form $y' = -ky,$ where $k > 0$ sets the time scale and the half-life is $\ln(2)/k.$

Ordinary and partial differential equations

7.An ordinary differential equation has one independent variable, and all its derivatives are taken with respect to that variable. If $y$ depends on $x,$ the equation below is ordinary because $x$ is its only independent variable:

$$y'' + x^2y' - y = \sin(x)$$

8.A partial differential equation has an unknown function of two or more independent variables and at least one partial derivative. If $u = u(x,t)$ is the temperature at position $x$ and time $t,$ the one-dimensional heat equation is:

$$\frac{\partial u}{\partial t} = k\frac{\partial^2u}{\partial x^2} \qquad k > 0$$

9.The derivative with respect to $t$ gives the rate of change of temperature in time, while the second derivative with respect to $x$ measures how the spatial temperature gradient changes with position.

Classification by order

10.The order of an equation is the highest derivative order that occurs.

12.For a scalar equation solved for its highest derivative, standard initial data prescribe one value for a first-order equation and two for a second-order equation. Under suitable existence and uniqueness hypotheses, these data determine a unique local solution.

13.The general solution of the second-order equation $y'' + 4y = 0$ is the following linear combination of sine and cosine:

$$y(x) = A\cos(2x) + B\sin(2x)$$

14.The values of $y(x_0)$ and $y'(x_0)$ fix the two constants. At $x_0 = 0,$ for example, $y(0) = A$ and $y'(0) = 2B.$

First-order differential equations

15.In a first-order differential equation, $y'$ occurs but higher-order derivatives do not. Its implicit form is:

$$F(x,y,y') = 0$$

16.If the equation can be solved for $y',$ its normal form is:

$$y' = f(x,y)$$

17.A first-order equation is separable if it can be written as $y' = g(x)h(y),$ and it is linear if it has the form $y' + p(x)y = q(x).$ Some first-order equations are both separable and linear. One separable equation is:

$$y' = 2xy$$

18.Assume that $y \neq 0$ on an interval. Dividing by $y$ separates the two variables:

$$\frac{1}{y}\frac{dy}{dx} = 2x$$

19.Since $\frac{d}{dx}\ln|y(x)| = \frac{y'(x)}{y(x)},$ integrating both sides with respect to $x$ gives:

$$\ln|y| = x^2 + C_1$$

20.The absolute value inside the logarithm is needed because $\ln|y|$ is an antiderivative of $1/y$ on both $(-\infty,0)$ and $(0,\infty).$ Exponentiation gives $|y| = e^{C_1}e^{x^2}.$ Since $y$ does not vanish on the interval, it has a constant sign, and we set $C = \pm e^{C_1}.$ The solution family is:

$$y(x) = Ce^{x^2}$$

21.Division by $y$ excluded the zero solution, so we include $C = 0$ in the family. The chain rule verifies the formula:

$$y'(x) = 2xCe^{x^2} = 2xy(x)$$

22.This equation is also linear because it can be written as $y' - 2xy = 0,$ whose coefficient $-2x$ depends only on the independent variable. Order and linearity are independent. Equations of order one include linear and nonlinear cases, while linear equations occur at every order.

Linear and nonlinear differential equations

23.A linear differential equation of order $n$ has the form:

$$a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \cdots + a_1(x)y' + a_0(x)y = g(x)$$

24.Each coefficient is a function of the independent variable $x$ alone, and the leading coefficient $a_n(x)$ does not vanish on the interval under consideration. The unknown function and its derivatives occur only to the first power, no two of them are multiplied together, and none is the argument of a nonlinear function. The equation is homogeneous when $g(x) = 0$ throughout the interval and nonhomogeneous otherwise.

25.Write the left-hand side as $L[y].$ If $y_1$ and $y_2$ are solutions of the homogeneous equation and $c_1,c_2$ are constants, linearity gives:

$$L[c_1y_1 + c_2y_2] = c_1L[y_1] + c_2L[y_2] = 0$$

26.The set of homogeneous solutions is therefore a vector space. When the coefficients are continuous and $a_n$ does not vanish on the interval, this vector space has dimension $n.$ Any $n$ linearly independent homogeneous solutions are a basis, so every homogeneous solution is their linear combination. The family $A\cos(2x) + B\sin(2x)$ above is the general solution of $y'' + 4y = 0.$

27.The difference of two solutions of a nonhomogeneous equation is a solution of the associated homogeneous equation because subtracting their equations cancels the right-hand side $g(x).$ Every solution therefore has the form $y = y_p + y_h,$ where $y_p$ is one fixed solution of the nonhomogeneous equation and $y_h$ is any homogeneous solution. These properties follow from linearity and do not hold in general for nonlinear equations.

28.When $n = 1,$ division by the leading coefficient gives:

$$y' + p(x)y = q(x)$$

29.The equation below is linear and first-order:

$$y' + 2y = e^x$$

30.To turn the left-hand side into a product derivative, choose an integrating factor $\mu(x)$ such that $\mu y' + 2\mu y = (\mu y)'.$ By the product rule, this requires $\mu' = 2\mu,$ so we take $\mu(x) = e^{2x}.$ Multiplication by $e^{2x}$ gives:

$$e^{2x}y' + 2e^{2x}y = e^{3x}$$

31.The equation is therefore equivalent to:

$$\left(e^{2x}y\right)' = e^{3x}$$

32.Integration and division by $e^{2x}$ give:

$$ \begin{align} e^{2x}y &= \frac{1}{3}e^{3x} + C \\[6pt] y &= \frac{1}{3}e^x + Ce^{-2x} \end{align} $$

33.The term $\frac{1}{3}e^x$ is a particular solution, and $Ce^{-2x}$ is the general solution of the associated homogeneous equation.

34.The general integrating-factor method is developed in first-order linear differential equations.

35.In a nonlinear differential equation, the unknown function or at least one of its derivatives appears in a nonlinear term. The examples below are nonlinear for different reasons:

37.For example, $y' = x+y^2$ has order one and is nonlinear, whereas $y'' + 4y = 0$ has order two and is linear.

Autonomous equations and systems

38.In an autonomous ordinary differential equation, the independent variable does not occur explicitly. A first-order autonomous equation is of the form $y' = f(y).$ The logistic equation for a population $P(t)$ is autonomous:

$$P' = rP\left(1-\frac{P}{K}\right) \qquad r,K > 0$$

39.The right-hand side depends only on the current population. It vanishes at $P = 0$ and $P = K,$ so the constant functions $P(t) = 0$ and $P(t) = K$ are equilibrium solutions. The equation is also first-order, separable, and nonlinear.

40.If $0 < P(0) < K,$ the solution remains between the two equilibrium values. In this range the right-hand side is positive, so $P' > 0$ and the solution is increasing. Differentiating the equation gives $P'' = rP'(1-2P/K),$ so the concavity changes when $P = K/2.$ The solution is therefore sigmoidal.

41.A system of differential equations is a collection of equations for several unknown functions and their derivatives. A model for two interacting populations is:

$$ \begin{align} x' &= ax-bxy \\[6pt] y' &= -cy+dxy \end{align} $$

42.Here $x$ and $y$ are unknown functions of time, and $a,b,c,d$ are positive constants. The products $xy$ make the system nonlinear. If an equation of order $n$ is solved for $y^{(n)},$ set $z_1 = y,$ $z_2 = y',$ $\ldots,$ and $z_n = y^{(n-1)}.$ Then $z_1' = z_2,\ldots,z_{n-1}' = z_n,$ and the original equation determines $z_n',$ so the scalar equation becomes a first-order system.

Initial and boundary conditions

43.Initial and boundary conditions restrict the family of solutions of a differential equation. An initial-value problem prescribes all its data at the same point of the independent variable. The problem below prescribes one value at $x = 0:$

$$y' = 2y \qquad y(0) = 3$$

44.The general solution of the differential equation is $y(x) = Ce^{2x}.$ The condition $y(0) = 3$ gives $C = 3,$ so the solution is:

$$y(x) = 3e^{2x}$$

45.For a scalar equation in normal form of order $n,$ the initial data are the values of $y,y',\ldots,y^{(n-1)}$ at one point.

46.A boundary-value problem prescribes conditions at different points, often at the endpoints of an interval. The problem below prescribes values at two points:

$$y'' + y = 0 \qquad y(0) = 0 \qquad y\left(\frac{\pi}{2}\right) = 1$$

47.The general solution of the differential equation is $y(x) = A\cos(x) + B\sin(x).$ The first boundary condition gives $A = 0,$ and the second then gives $B = 1.$ The solution is therefore:

$$y(x) = \sin(x)$$

48.Conditions alone do not ensure a unique solution. The equation $y'' + y = 0$ on $[0,\pi]$ has no solution with $y(0) = 0$ and $y(\pi) = 1,$ since every function $B\sin(x)$ vanishes at both endpoints. With the conditions $y(0) = y(\pi) = 0,$ every function $B\sin(x)$ is a solution. Thus a boundary-value problem may have no solution, one solution, or several solutions.

49.An existence theorem states sufficient conditions for at least one solution, while a uniqueness theorem states sufficient conditions under which two solutions with the same data are equal. Continuity alone does not guarantee uniqueness. The function $f(y) = 3\sqrt[3]{y^2}$ is continuous but not locally Lipschitz at $y = 0.$ The problem $y' = f(y),$ $y(0) = 0,$ has both $y(x) = 0$ and $y(x) = x^3$ as solutions. For $y(x) = x^3,$ both sides of the equation are $3x^2,$ while both sides vanish for $y(x) = 0.$

50.For a linear equation, suppose that its coefficients and right-hand side are continuous on an interval $I$ and that its leading coefficient does not vanish there. An initial-value problem posed at a point of $I$ then has exactly one solution on the whole interval $I.$ For a nonlinear equation $y' = f(x,y),$ continuity of $f$ and local Lipschitz continuity with respect to $y$ near the initial point give a unique local solution. The problem $y' = e^y,$ $y(0) = 0,$ has the solution $y(x) = -\ln(1-x)$ on the maximal interval $(-\infty,1).$ The right-hand side is defined for every real $y,$ but the solution is unbounded as $x$ approaches $1$ from the left. Thus a local solution need not extend to the whole domain of $f.$

Explicit, implicit, and numerical descriptions

51.A solution is explicit when the dependent variable is isolated, as in $y(x) = Ce^{-x}.$ A solution is implicit when it is given by a relation between the variables. For $yy' = x,$ an implicit solution family is:

$$y^2-x^2 = C$$

52.Differentiation with respect to $x$ gives $2yy'-2x = 0,$ which is equivalent to the differential equation. Away from points where $y = 0,$ the explicit branches are $y = \pm\sqrt{x^2+C}.$ Each branch is a solution on an interval where $x^2+C > 0.$

53.The solution of an initial-value problem may not be expressible in elementary functions. For example, by the Fundamental Theorem of Calculus, the problem $y' = e^{-x^2},$ $y(0) = 0,$ has the solution $y(x) = \int_0^x e^{-t^2} \ dt,$ which is not an elementary function. When no elementary formula is available, one can establish existence and uniqueness, determine qualitative properties such as monotonicity and stability, or compute numerical approximations. For the initial-value problem $y' = f(x,y),$ $y(x_0) = y_0,$ Euler's method uses the tangent line at the current point to approximate the solution over one step. With step size $h,$ the approximation $y(x_n+h) \approx y(x_n) + hy'(x_n)$ and the differential equation give the recurrence:

$$x_{n+1} = x_n + h, \qquad y_{n+1} = y_n + hf(x_n,y_n)$$

54.The values $y_n$ approximate the solution at the points $x_n.$ If a linear equation has analytic coefficients and an analytic right-hand side near a point where its leading coefficient is nonzero, each local solution has a convergent power-series expansion there.