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Unformatted text preview: Nonhomogeneous Linear Equations In this section we learn how to solve second-order nonhomogeneous linear differential equa- tions with constant coefficients, that is, equations of the form where , , and are constants and is a continuous function. The related homogeneous equation is called the complementary equation and plays an important role in the solution of the original nonhomogeneous equation (1). Theorem The general solution of the nonhomogeneous differential equation (1) can be written as where is a particular solution of Equation 1 and is the general solution of the complementary Equation 2. Proof All we have to do is verify that if is any solution of Equation 1, then is a solution of the complementary Equation 2. Indeed We know from Additional Topics: Second-Order Linear Differential Equations how to solve the complementary equation. (Recall that the solution is , where and are linearly independent solutions of Equation 2.) Therefore, Theorem 3 says that we know the general solution of the nonhomogeneous equation as soon as we know a par- ticular solution . There are two methods for finding a particular solution: The method of undetermined coefficients is straightforward but works only for a restricted class of func- tions . The method of variation of parameters works for every function but i0s usually more difficult to apply in practice. The Method of Undetermined Coefficients We first illustrate the method of undetermined coefficients for the equation where ) is a polynomial. It is reasonable to guess that there is a particular solution that is a polynomial of the same degree as because if is a polynomial, then is also a polynomial. We therefore substitute a polynomial (of the same degree as ) into the differential equation and determine the coefficients. EXAMPLE 1 Solve the equation . SOLUTION The auxiliary equation of is r 2 r 2 r 1 r 2 y y 2 y y y 2 y x 2 G y p x ay by cy y G y p G x ay by cy G x G G y p y 2 y 1 y c c 1 y 1 c 2 y 2 t x t x ay by cy ay p by p cy p a y y p b y y p c y y p ay ay p by by p cy cy p y y p y y c y p y x y p x y c x 3 ay by cy 2 G c b a ay by cy G x 1 1 with roots , . So the solution of the complementary equation is Since is a polynomial of degree 2, we seek a particular solution of the form Then and so, substituting into the given differential equation, we have or Polynomials are equal when their coefficients are equal. Thus The solution of this system of equations is A particular solution is therefore and, by Theorem 3, the general solution is If (the right side of Equation 1) is of the form , where and are constants, then we take as a trial solution a function of the same form, , because the derivatives of are constant multiples of ....
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