FIRST-ORDER DIFFERENTIAL EQUATIONS

MIT OpenCourseWarehttp://ocw.mit.edu 18.034 Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. � � � � UNIT I: FIRST-ORDER DIFFERENTIAL EQUATIONS We set forth fundamental principles in the analysis of differential equations. We illustrate the use of integration to find the solutions of first-order linear differential equations and special classes of first-order nonlinear differential equations, called separable equations. Substitution techniques are used in studying linear fractional equations and special kind of second-order differential equatioons LECTURE 1. INTEGRATION AND SOLUTIONS We recall the fundamental theorem of calculus d x (1.1) f(s)ds = f(x),dx x0 if f is continuous on an interval x0 ∈ I. A solution of the differential equation (1.2) dy = f(x)dx is the function y = φ(x) which satisfies the differential equation on I. Upon inspection of (1.1), x then, y = f(s)ds is a solution of (1.2). This leads to an existence result. x0 Theorem 1.1. If f(x) is continuous on an internval xo ∈ I then given an arbitrary number y0 there exists a unique solution of (1.2) satisfying y(x0)= y0. The solution is given as x y(x)= y0 + f(s)ds. x0 Exercise. Prove the uniqueness. Remark. 1. The theorem specifies the interval of existence (x0 ∈ I) and the class of functions considered (the class of continuous functions). It asserts the existence and uniqueness of a solution, prescribed the initial condition y(x0)= y0. 2. In the statement of the theorem, the interval of existence is I, regardless of the initial conditiion It is a special property of linear equations. For nonlinear equations, in general, the interval of existence depends on the initial value, e.g. the solution of the initial value problem dy 2= y, y(0) = y0 =0,dx �1is given as y(x)= (1/y0) − x . It is defined on x ∈ [0, 1/y0) for y0 > 0. x 3. The definite integral f(s)ds is defined as a limit of Riemann sums, as long as f is continuxx � ous; it doesn’t need to find a formal expression for the indefinite integral f(s)ds to give meaning � x 2to the definite integral, e.g. the error function erf(x)= e−sds and the sine integral function� 0 Si(x)= 0 x(sin s)/sds are commonly defined as definite integrals. 1 � � As an illustration, the solution of the initial value problem dy 2= sin x, y(0) = 0 dx is given by the Fresnel sine integral function S(x)= 0 x sin s2 ds. There is no elementary function F such that F �(x) = sin x2, but the function S(x) defined as a definite integral gives a perfectly good function. The preceding discussion leads to how to solve differential equations of the form (1.2) by in-x spection. For any x0, one solution is the function f(s)ds. Other solutions are, then, obtained x0 by adding an arbitrary constant C to this “particular” solution. Thus, the solutions of y� = e−x2 are the functions y = � e−s2 ds =(√π/2)erf(x)+ C. From any one solution curve of (1.2), the others are obtained by the vertical translations (x,y) �→ (x,y + C) and they form a one-parameter family of curves, one for each value of the parameter C. Quadrature. When the solution of a differential equation is expressed by a formula involving one or more integrations, it is said that the equation is solvable by quadrature. The term “quadrature” has its historical origin in the connection of integration with area. In plane geometry, a problem of quadrature, such as quadrature of the circle is a problem about the area of a plane figure. Not all differential equations can be solved by quadrature. In the following lecture, we will show that the first-order linear equation y� + p(x)y = q(x) can be solved by quadrature. But, the second-order differential equation y�� + p(x)y� + q(x)y = r(x) cannot be solved, in general, by quadrature, except for some special cases. The next simples type of differential equation is (1.3) dy = g(y). dx Such a differential equation is invariant under horizontal translations (x,y) �→ (x + c,y). Geometricaally it means that any horizontal line is cut by all solution curves at the same angle (such lines are called “isoclines”). Therefore, if y = φ(x) is a solution of (1.3), then so is y = φ(x + c) for any c. The differential equation (1.3) can be solved by writing it as dy/g(y)= dx and integrating. Example 1.2. Consider (1.4) dy = y 2 − 1. dx Since y2 − 1=(y − 1)(y + 1), the constant functions y = ±1 are particular solutions of (1.4). They are called steady states, stationary solutions or equilibria, in the sense that these solutions are independent of x. Next, if y< 1 then y2 − 1 < 0, and follows dy/dx < 0. That is, the solution curve is decreasing. || 2On the other hand, if |y| > 1, then dy/dx = y− 1 > 0, and the solution curve is increasing. It gives us the qualitative behavior of solutions curves of (1.4). Lecture 1 2 18.034 Spring 2009 � � Figure 1.1. Qualitative behavior of solutions of y� = y2 − 1. Using the partial fractions and separating variables (we will discuss this technique in detail later), (1.4) is written as ��112dx = dy y − 1 − y +1 . Then, by integration, we obtain y(x)= 1 ± e2(x−c) = tanh (c − x).1 � e2(x−c) coth This procedure of separating variables “loses” the particular solutions y = ±1, but it gives all other solutions. Note that if y = φ(x) is a solution of (1.4) then so is 1/y =1/φ(x). Exercise. Discuss y� = y3 − y. Lecture 1 3 18.034 Spring 2009