Honors Differential Equations
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1 MIT OpenCourseWare Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit:
2 LECTURE 13. INHOMOGENEOUS EQUATIONS We discuss various techniques for solving inhomogeneous linear differential equations. Variation of parameters: the Lagrange procedure. Let us consider the linear second-order differential operator (13.1 Ly = y + p(ty + q(ty with variable coefficients. If a nonvanishing solution of a homogeneous equation Ly = 0 is known, then the corresponding inhomogeneous equation Ly = f can be solved, in general, by two integrations. It was discovered by Lagrange that if two linearly independent solutions of Ly = 0 are known, then the inhomogeneous equation Ly = f can be solved by a single integration. Let u and v be a pair of linearly independent solutions of Ly = 0, and form the expression (13.2 y = au + bv. If a and b are constant, this represents the general solution of Ly = 0. We will the inhomogeneous equation Ly = f by choosing a trial solution of this form, but with a and b functions of t, rather than constants. The method is called the method of variation of parameters. Let a and b are differentiable functions of t. By differentiation, We require y = (au + bv + (a u + b v. (13.3 a u + b v = 0 so that y = au + bv. This simplifies the calculation of the second derivative, and Therefore, y = (au + bv + (a u + b v. Ly = y + py + qy = alu + blv + a u + b v = a u + b v. The second equality uses that Lu = Lv = 0. Solving Ly = f in the form in (13.2 then reduces to the linear system a u + b v = 0, a u + b v = f, in the unknown a and b. In the matrix form, ( ( ( u v a 0 =. u v b f By Cramers rule, we solve the system, and 0 v u 0 u f a f v =, b =. u v u v u v u v 1
3 Here, the notation stands for the determinant of the matrix. The denominator is the Wromskian W (u, v, so that we may write them as a fv =, b fu =. W (u, v W (u, v Finally, by integration, we obtain the Lagange formula fv fu (13.4 y(t = u(t dt + v(t dt. W (u, v W (u, v Lagrange s procedure extends to equations of order n and it represents an important advance in the theory of differential equations. A similar idea already appeared. For example, when studying the linear first-order differential equations, we replaced the homogeneous solution ce P by ve P, where v is a function. Example Consider the Euler equation (13.5 x 2 y 2xy + 2y = x 2 f(x, x > 0, where the prime denotes the differentiation in the x-variable. By the technique discussed in the previous lecture, we compute u = x, v = x 2, W (u, v = x 2. For x > 0, we write (13.5 as 2 2 y y + y = f(x. x x 2 Then Lagrange s formula (13.4 gives 2 f(x y(x = x f(xdx + x x 2. Exercise. If f(x = x m, where m is a constant, in the above example, show that a particular solution of (13.5 is x log x if m = 1, y p (x = x 2 log x if m = 0, x m+2 otherwise. m(m + 1 The general solution of (13.5 is y(t = c 1 x + c 2 x 2 + y p (x. The Green s function: initial value problems. As an important application of the formula (13.4 we can find an integral representation of the initial value problem for Ly = f, where L is given in (13.1. Let t 0 be a point on the interval I. Integrating (13.4 from t 0 to t, t f(t v(t t f(t u(t y(t =u(t dt + v(t dt t W (t 0 t 0 W (t t u(t v(t u(tv(t = f(t dt u(t v (t u (t v(t This function satisfies the conditions t 0 y(t 0 = 0, y (t 0 = 0. Lecture Spring 2009
4 Indeed, y(t = a(tu(t + b(tv(t and y (t = a(tu (t + b(tv (t where a(t = f (t v(t dt W ( t and f(t u(t b(t = dt. W (t In summary, the function defined as t G(t, t f(t dt, G(t, t u(t v(t u(tv(t y(t = = t u(t v (t u (t v(t 0 solves the initial value problem, Ly = f, y(t 0 = 0, y (t 0 = 0. The function G(t, t is called the Green s function. Example We continue studying the Euler equation (13.5 satisfying the initial conditions y(x 0 = 0, y (x 0 = 0 for some x 0 > 0. The solution has an integral representation x f(t y(x = x (x t dt. t For example, if f(x = x sin x then x y(x = x (x t sin tdt = x(x x 0 cos x 0 x(sin x sin x 0. x 0 x 0 Exercise. (The Green s function: boundary value problem We consider the boundary value problem y + p(ty + q(ty = f(t on (t 1, t 2, y(t 1 = y(t 2 = 0. If u and v are linearly independent solutions of the homogeneous equation y + py + qy = 0, then show that the solution of the boundary value problem is given by t 2 y(t = G(t, tf(t dt, t 1 u(t v(t W (t if t 1 t t, where G(t, t = u(tv(t W (t if t t t 2. The method of annihilators. We introduce another method of finding a particular solution of linear inhomogeneous differential equation with constant coefficients. Let Ly = y (n + p 1 y (n p n 1 y + p n y, where p j are real constants. We study the differential equation Ly = f, where f is a sum of functions of type t r e λt, t r e µt sin νt, t r e µt cos νt. Note that these functions arise as basis solutions of linear homogeneous differential equations with constant coefficients. We find a differential operator A satisfies Af = 0, then we reduce solving Ly = f to solving the homogeneous equation LAy = 0. Such an operator A is called an annihilator of f. We illustrate with an example. Lecture Spring 2009
5 Example We consider the differential equation (13.6 y 5y 6y = te t. Let L = D 2 5D 6 = (D 2(D 3. Then (13.6 is written as Ly = te t. By the exponential shift law for D, we recognize that te t is a solution of the differential equation (D 1 2 y = 0. In other words, (D 1 2 is an annihilator of te t. Applying (D 1 2 in (13.6, we obtain the homogeneous differential equation (D 2(D 3(D 1 2 y = 0. It is easy to see that e t, te t, e 2t, e 3t form a basis of solutions of the above equation. Hence, we set a solution of (13.6 as y(t = c 1 e t + c 2 te t + c 3 e 2t + c 4 e 3t, and determine the constants c j. Since Le 2t = 0 and Le 3t = 0, moreover, we may set c 3 = c 4 = 0. Hence, y(t = c 1 e t + c 2 te t. We compute Ly = (D 2 5D 6(c 1 e t + c 2 te t = (2c 1 3c 2 e t + 2c 2 te t = te t to obtain c 1 = 3/4 and c 2 = 1/2. Therefore, a particular solution of (13.6 is y(t = 3/4e t + 1/2te t. Lecture Spring 2009
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