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Unformatted text preview: McGill University Math 263: Differential Equations for Engineers CHAPTER 3: NTH ORDER DIFFERENTIAL EQUATIONS (III) 1 Finding a Particular Solution for Inho mogeneous Equation In this lecture we shall discuss the methods for producing a particular solution of a special kind for the general linear DE. In what follows we shall use the operator method to find a particular solution of inhomogeneous equations. 1.1 The Annihilator and the Method of Undeter mined Constants We consider the linear equation with constant coefficients first. Given L ( y ) = P ( D )( y ) = ( a D ( n ) + a 1 D ( n 1) + + a n D ) y = b ( x ) . Assume that the inhomogeneous term b ( x ) is a solution of lin ear equation: Q ( D ) ( b ( x ) ) = 0 . 00 Then we can transform the original inhomogeneous equa tion to the homogeneous equation by applying the differential operator Q ( D ) to its both sides, ( D )( y ) = Q ( D ) P ( D )( y ) = 0 . The operator Q ( D ) is called the Annihilator . The above method is also called the Annihilator Method . Example 1. Solve the initial value problem y 000 3 y 00 + 7 y 5 y = x + e x , y (0) = 1 ,y (0) = y 00 (0) = 0 . This DE is nonhomogeneous. The associated homoge neous equation was solved in the previous lecture. Note that in this example, In the inhomogeneous term b ( x ) = x + e x is in the kernel of Q ( D ) = D 2 ( D 1) . Hence, we have D 2 ( D 1) 2 ( D 1) 2 + 4 / ( y ) = 0 which yields y = Ax + B + Cxe x + c 1 e x + c 2 e x cos(2 x )+ c 3 e x sin(2 x ) . This shows that there is a particular solution of the form y P = Ax + B + Cxe x (1) 01 which is obtained by discarding the terms: c 1 e x + c 2 e x cos(2 x ) + c 3 e x sin(2 x ) in the solution space of the associated homogeneous DE. Now the remaining part of problem is to determine the arbitrary constants A,B,C in the form of particular solution (1). In doing so, substituting (1) in the original DE we get y 000 3 y 00 + 7 y 5 y = 7 A 5 B 5 Ax Ce x which is equal to x + e x if and only if 7 A 5 B = 0 , 5 A = 1 , C = 1 . So that A = 1 / 5 , B = 7 / 25 , C = 1 . Hence the general solution is y = c 1 e x + c 2 e x cos(2 x )+ c 3 e x sin(2 x ) x/ 5 7 / 25 xe x . To satisfy the initial condition y (0) = 0 ,y (0) = y 00 (0) = 0 we must have c 1 + c 2 = 32 / 25 , c 1 + c 2 + 2 c 3 = 6 / 5 , c 1 3 c 2 + 4 c 3 = 2 (2) 02 which has the solution c 1 = 3 / 2 ,c 2 = 11 / 50 ,c 3 = 1 / 25 . It is evident that if the function b ( x ) can not be annihilated by any linear operator Q ( D ) , the annihilator method will not applicable. 1.2 The Annihilators for Some Types of Func tions The annihilators Q ( D ) for some types of functions F ( x ) are as follows: For the functions F ( x ) = a + a 1 x + + a k x k , we have the annihilator: Q ( D ) = D k +1 ....
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This note was uploaded on 12/01/2010 for the course MATH 263 taught by Professor Sidneytrudeau during the Fall '09 term at McGill.
 Fall '09
 SidneyTrudeau
 Math, Differential Equations, Equations

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