Chapter 3 Lecture 8

Chapter 3 Lecture 8 - McGill University Math 263...

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McGill University Math 263: Differential Equations for Engineers CHAPTER 3: N-TH 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 0 D ( n ) + a 1 D ( n - 1) + · · · + a n D 0 ) y = b ( x ) . Assume that the inhomogeneous term b ( x ) is a solution of lin- ear equation: Q ( D ) ( b ( x ) ) = 0 . 0-0

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Then we can transform the original inhomogeneous equa- tion to the homogeneous equation by applying the differential operator Q ( D ) to its both sides, Φ( 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 0 - 5 y = x + e x , y (0) = 1 , y 0 (0) = y 00 (0) = 0 . This DE is non-homogeneous. 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) 0-1
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 0 - 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 (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) 0-2

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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 0 + a 1 x + · · · + a k x k , we have the annihilator: Q ( D ) = D k +1 . For the functions F ( x ) = ( a 0 + a 1 x + · · · + a k x k )e ax , we have the annihilator: Q ( D ) = ( D - a ) k +1 . 0-3
For the functions F ( x ) = e λx h a 0 cos μx + b 0 sin μx i , we have the annihilator: Q ( D ) = £ ( D - λ ) 2 + μ 2 / . For the functions F ( x ) = ( a 0 + a 1 x + · · · + a k x k )e λx cos μx, or F ( x ) = ( b 0 + b 1 x + · · · + b k x k )e λx sin μx, we have the annihilator: Q ( D ) = £ ( D - λ ) 2 + μ 2 / k +1 .

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