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.� � � � ���� ���� ���� ���� ���� ���� ���� ���� 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 differ-ential operator (13.1) Ly = y�� + p(t)y� + q(t)y 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 inhomoge-neous 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, y� = (au� + bv�) + (a�u + b�v). We require (13.3) a�u + b�v = 0 so that y� = au� + bv�. This simplifies the calculation of the second derivative, and y�� = (au�� + bv��) + (a�u� + b�v�). Therefore, 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 0 u u� f a� = f v� u v u� v� , b� = u v u� v� . 1� � � � � � � 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 (13.4) y(t) = u(t) −fv dt + v(t) fu 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 ceP by veP , where v is a function. Example 13.1. Consider the Euler equation (13.5) x 2 y�� − 2xy� + 2y = x 2f(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).2x xThen Lagrange’s formula (13.4) gives f(x) 2x. y(x) = −x f(x)dx + x 2 Exercise. If f(x) = xm, where m is a constant, in the above example, show that a particular solution of (13.5) is yp(x) = ⎧⎪⎪⎪⎨ ⎪⎪⎪⎩ −x log x if m = −1, x2 log x if m = 0, m+2xotherwise. m(m + 1) The general solution of (13.5) is y(t) = c1x + c2x2 + yp(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 t0 be a point on the interval I. Integrating (13.4) from t0 to t, t −f(t�)v(t�) t f(t�)u(t�) y(t) =u(t) dt� + v(t) dt� W (t�)t0 t0 W (t�) t u(t�)v(t) − u(t)v(t�) f(t�)dt�= u(t�)v�(t�) − u�(t�)v(t�)t0 This function satisfies the conditions y(t0) = 0, y�(t0) = 0. Lecture 13 2 18.034 Spring 2009� � � � Indeed, y(t) = a(t)u(t) + b(t)v(t) and y�(t) = a(t)u�(t) + b(t)v�(t) where a(t) = −fW (t�()tv�)(t�)dt� 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(t)v(t�) y(t) = u(t�)v�(t�) − u�(t�)v(t�)t0 solves the initial value problem, Ly = f, y(t0) = 0, y�(t0) = 0. The function G(t, t�) is called the Green’s function. Example 13.2. We continue studying the Euler equation (13.5) satisfying the initial conditions y(x0) = 0, y�(x0) = 0 for some x0 > 0. The solution has an integral representation x f(t) y(x) = x (x − t) dt. x0 For example, if f(x) = x sin x then t x y(x) = x (x − t) sin tdt = x(x − x0) cos x0 − x(sin x − sin x0). x0 Exercise. (The Green’s function: boundary value problem) We consider the boundary value prob-lem y�� + p(t)y� + q(t)y = f(t) on (t1, t2), y(t1) = y(t2) = 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 t2 y(t) = t1 G(t�, t)f(t�)dt�, u(t�)v(t) W (t�) u(t)v(t�) W (t�) if t1 � t� � t, if t � t� � t2. where G(t�, t) = ⎧ ⎪⎪⎨ ⎪⎪⎩ 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) + p1y(n−1) + + pn−1y� + pny,· · · where pj are real constants. We study the differential equation Ly = f, where f is a sum of functions of type tr e λt , tr eµt sin νt, tr eµt 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 13 3 18.034 Spring 2009Example 13.3. We consider the differential equation (13.6) y�� − 5y� − 6y = tet . Let L = D2 − 5D − 6 = (D − 2)(D − 3). Then (13.6) is written as Ly = tet . By the exponential shift law for D, we recognize that tet is a solution of the differential equation (D − 1)2y = 0. In other words, (D − 1)2 is an annihilator of tet. Applying (D − 1)2 in (13.6), we obtain the homogeneous differential equation (D
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