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 II: SECOND-ORDER LINEAR EQUATIONS We discuss the existence and uniqueness for second-order linear differential equations with constant coefficients by means of general principles which are valid for equations with variable coefficients. We develop techniques of the Wronskian and apply to study the oscillatory behav-ior. We also give qualitative results which depend on conditions at an interior maximum and minimum and apply to second-order equations. LECTURE 6. SECOND-ORDER LINEAR EQUATIONS The most intensively studied ordinary differential equations are second-order linear equations of the form (6.1) p0(t)y�� + p1(t)y� + p2(t)y = f(t), dwhere � = is differentiation in the t-variable, and pj (t), f(t), j = 0, 1, 2, are continuous functions dt on an interval I. Remark 6.1 (Remark on singular points). If p0(t) vanishes at some point, say t0, then we say that (6.1) has a singular point at t0. For example, the Legendre equation ((1 − t2)y�)� + λy = 0 or (1 − t2)y�� − 2ty� + λy = 0 has singular points at t = ±1. While the equation admits polynomial solutions, called the Le-gendre polynomials, for λ = n(n + 1), other nontrivial solutions develop a singularity at either t = 1 or t = −1. We do not treat singular points in this course. Bases of solutions. If p0(t) = 0 � on I then (6.1) reduces to the normal form (6.2) y�� + p(t)y� + q(t)y = f(t), where p(t), q(t) and f(t) are continuous on I. We consider the corresponding homogeneous equation (6.3) y�� + p(t)y� + q(t)y = 0. Since the equation is linear, the principle of superposition applies, and if φ and ψ are solutions of (6.3) then so is their linear combination c1φ + c2ψ for any constants c1, c2. Definition 6.2. On a given interval two functions are said linearly dependent if one of them is a constant multiple of the other. They are said linearly independent, otherwise. For example, e−t and e−2t are linearly independent on every interval of t, although they agree at t = 0. If φ and ψ are solutions of (6.3) and, in addition, if φ and ψ are linearly independent then every solution of (6.3) is of the form c1φ + c2ψ for some constants c1, c2. We call such a pair of solutions {φ, ψ} a basis of solutions of (6.3). The existence of a basis of solutions is a fundamental result for (6.3) and for a broader class of linear homogeneous differential equations, and we will prove it later. 1Constant-coefficient differential equations. We construct a basis of solution of a second-order constant-coefficient differential equation (6.4) y�� + py� + qy = 0, where p, q are constants. The trick is to set y(t) = eλt, where λ ∈ C. We use (eλt)� = λeλt and (eλt)�� = λ2eλt to write (6.4) as e λt(λ2 + pλ + q) = 0. Since eλt = 0� , it follows that y(t) = eλt is a solution of (6.4) if and only if λ is a root of the quadratic polynomial (6.5) p(λ) = λ2 + pλ + q, called the characteristic polynomial of (6.4). The characteristic polynomial has two roots (counting multiplicity) √Δ 2λ = −p ±2 , Δ = p − 4q. We divide our discussion according to the nature of the roots, which is determined by the sign of the discriminant Δ. If Δ > 0 then both roots are real, and e(−p+√Δ)t/2 , e(−p−√Δ)t/2 are two solutions of (6.4). Moreover, they are linearly independent. Therefore, they form a basis of solutions of (6.4). If Δ = 0 then the method yields only the single solution e−pt/2. Using the method suggested by Lagrange, we try ve−pt/2 as a second solution, where v = v(t). Upon substitution, (6.4) becomes (ve−pt/2)�� + p(ve−pt/2)� + qve−pt/2 = v��e−pt/2 = 0. Thus, v�� = 0 must hold and we obtain the family of solutions y(t) = (c1 + c2t)e−pt/2. Since e−pt/2 , te−pt/2 are linearly independent, they form a basis of solutions of (6.4). If Δ < 0, then both roots are complex, and e−pt/2+i√−Δt/2 , e−pt/2−i√−Δt/2 are two (complex) solutions of (6.4). Then, we infer that u + v = e−pt/2 cos(√−Δt/2),u − v = e−pt/2 sin(√−Δt/2)2 2i are two real solutions of (6.4). Since they are linearly independent, they form a basis of (real) solutions of (6.4). The principle of equating real parts holds for, more generally, variable-coefficient equations, and if z(t) = x(t) + iy(t) is a complex solution of z�� + p(t)z� + q(t)z = 0, where p(t) and q(t) are real-valued functions then so are x and y. Exercise. Prove the principle of equating real parts. Lecture 6 2 18.034 Spring 2009Inhomogeneous equations and initial value problems. Next, we turn to the inhomogeneous equation (6.2) with a general f(t). The following result enables us to get a family of solutions of (6.2) from a particular solution. Principle of the Complementary Solution. If yp is a solution of the inhomogeneous equation (6.2), called a particular solution and if φ and ψ form a basis of solutions of the homogeneous equation (6.3), then the general solution of (6.2) is given by yp + c1φ + c2ψ, where c1 and c2 are arbitrary constants. Proof. Let y be a solution of (6.2). Since y − yp satisfies the homogeneous equation (6.3), it should take the form c1φ + c2ψ for some constants c1, c2. � Example 6.3. Let us consider (6.6) y�� + y = 3 sin 2t. We try yp(t) = A sin 2t, where A is a constant, for a particular solution. Since y�� = −4A sin 2t, the differential equation becomes (−4A + A) sin 2t = 3 sin 2t. Thus, A = −1, and yp(t) = − sin 2t is a particular solution of (6.6). On the other hand, cos t and sin t form a basis of solutions of the corresponding homogeneous equation y�� + y = 0. Therefore, the general solution of (6.6) is given y(t) = − sin 2t + c1 cos t + c2 sin t. Particular solutions with polynomial forcing terms are treated similarly. As an illustration, we consider y�� + py� + qy = ct + d, where p, q, c, d are constants. We first look for a particular solution of the form yp(t) = at+b. Upon substitution, we obtain the equations qa = c, pa + qb = 0. Unless q = 0, these give the particular solution yp(t) = ct + qd − pc. q q2 If q = 0 but p = 0� , then we look for a particular solution as a quadratic polynomial; y�� + y� = x has a solution
View Full Document
Unlocking...