Chapter 3 Lecture 7

Chapter 3 Lecture 7 - McGill University Math 263:...

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Unformatted text preview: McGill University Math 263: Differential Equations for Engineers CHAPTER 3: N-TH ORDER DIFFERENTIAL EQUATIONS (II) 1 Solutions for Equations with Constants Coefficients In what follows, we shall first focus on the linear equations with constant coefficients: L ( y ) = a y ( n ) + a 1 y ( n- 1) + + a n y = b ( x ) and present two different approaches to solve them. 1.1 The Method with Undetermined Parameters To illustrate the idea, as a special case, let us first consider the 2-nd order Linear equation with the constant coefficients: L ( y ) = ay 00 + by + cy = f ( x ) . (1) The associate homogeneous equation is: L ( y ) = ay 00 + by + cy = 0 . (2) 0-0 1.2 Basic Equalities (I) We first give the following basic identities: D (e rx ) = r e rx ; D 2 (e rx ) = r 2 e rx ; D n (e rx ) = r n e rx . (3) To solve this equation, we assume that the solution is in the form y ( x ) = e rx , where r is a constant to be determined. Due to the properties of the exponential function e rx : y ( x ) = ry ( x ); y 00 ( x ) = r 2 y ( x ); y ( n ) = r n y ( x ) , (4) we can write L ( e rx ) = ( r )e rx . (5) for any given ( r,x ) , where ( r ) = ar 2 + br + c. is called the characteristic polynomial. From (5) it is seen that the function e rx satisfies the equation (1), namely L ( e rx ) = 0 , as long as the constant r is the root of the characteristic poly- nomial, i.e. ( r ) = 0 . 0-1 In general, the polynomial ( r ) has two roots ( r 1 ,r 2 ) : One can write ( r ) = ar 2 + br + c = a ( r- r 1 )( r- r 2 ) . Accordingly, the equation (16) has two solutions: ' y 1 ( x ) = e r 1 x ; y 2 ( x ) = e r 2 x . Three cases should be discussed separately. 1.3 Cases (I) ( r 1 > r 2 ) When b 2- 4 ac > , the polynomial ( r ) has two distinct real roots ( r 1 6 = r 2 ) . In this case, the two solutions, y ( x ); y 2 ( x ) are different. The following linear combination is not only solution, but also the general solution of the equation: y ( x ) = Ay 1 ( x ) + By 2 ( x ) , (6) where A,B are arbitrary constants. To prove that, we make use of the fundamental theorem which states that if y,z are two solutions such that y (0) = z (0) = y 0-2 and y (0) = z (0) = y then y = z . Let y be any solution and consider the linear equations in A,B Ay 1 (0) + By 2 (0) = y (0) , Ay 1 (0) + By 2 (0) = y (0) , (7) or A + B = y , Ar 1 + Br 2 = y . (8) Due to r 1 6 = r 2 , these conditions leads to the unique solution A,B . With this choice of A,B the solution z = Ay 1 + By 2 satisfies z (0) = y (0) , z (0) = y (0) and hence y = z . Thus, (6) contains all possible solutions of the equation, so, it is indeed the general solution. 1.4 Cases (II) ( r 1 = r 2 ) When b 2- 4 ac = 0 , the polynomial ( r ) has double root: r 1 = r 2 =- b 2 a . In this case, the solution y 1 ( x ) = y 2 ( x ) = e r 1 x ....
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Chapter 3 Lecture 7 - McGill University Math 263:...

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