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Unformatted text preview: Differential Equations & Linear Algebra Lecture 8 Jianli XIE Email: [email protected] Department of Mathematics, SJTU Fall 2010 Contents Method of variation of parameters An illustrative example Method of variation of parameters Contents Method of variation of parameters An illustrative example Method of variation of parameters Euler equations Contents Method of variation of parameters An illustrative example Method of variation of parameters Euler equations Reduction of order Variation of parameters The method of undetermined coefficients is limited to those equations for which we can easily write down the correct form of the particular solution in advance. The other method is known as variation of parameters . Its main advantage is that it is a general method : in principle at least, it can be applied to any equation, and it requires no detailed assumptions about the form of the solution. In fact, we can use this method to derive a formula for a particular solution of an arbitrary second order linear nonhomogeneous equation. An illustrative example Example Find a particular solution of y 00 + 4 y = 3 csc t. Solution The nonhomogeneous term involves a quotient of sin t , so the method of undetermined coefficients does not work. Note that the general solution of the corresponding homogeneous equation y 00 + 4 y = 0 is y c ( t ) = c 1 cos 2 t + c 2 sin 2 t . Now vary the constants c 1 and c 2 and assume y = u 1 ( t ) cos 2 t + u 2 ( t ) sin 2 t (1) is a solution of the nonhomogeneous equation. Before we substitute the expressions for y, y and y 00 into the equation so as to determine u 1 and u 2 , An illustrative example we do some analysis first. Without carrying out this substitution, we can anticipate that the result will be a single equation about u 1 , u 2 and their derivatives. Since there is only one equation and two unknown functions, we can impose a second condition about u 1 , u 2 and their derivatives, thereby obtaining two equations for the two unknown functions u 1 and u 2 . Now returning to the solution, from equation (1), we have y = 2 u 1 sin 2 t + 2 u 2 cos 2 t + u 1 cos 2 t + u 2 sin 2 t. An illustrative example Keeping in mind choosing a second condition on u 1 and u 2 , we require that u 1 cos 2 t + u 2 sin 2 t = 0 . (2) It then follows that...
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This note was uploaded on 05/17/2011 for the course ME 255 taught by Professor Jianlixie during the Fall '10 term at Shanghai Jiao Tong University.
 Fall '10
 JianliXie

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