Chapter 3 Lecture 7

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

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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 0 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 0 + cy = f ( x ) . (1) The associate homogeneous equation is: L ( y ) = ay 00 + by 0 + cy = 0 . (2) 0-0

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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 0 ( x ) = ry ( x ); y 00 ( x ) = r 2 y ( x ); · · · y ( n ) = r n y ( x ) , (4) we can write L ( e rx ) = φ ( r )e rx . (5) for any given ( r, x ) , where φ ( 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. φ ( r ) = 0 . 0-1
In general, the polynomial φ ( r ) has two roots ( r 1 , r 2 ) : One can write φ ( 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 > 0 , the polynomial φ ( 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 0-2

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and y 0 (0) = z 0 (0) = y 0 0 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 0 1 (0) + By 0 2 (0) = y 0 (0) , (7) or A + B = y 0 , Ar 1 + Br 2 = y 0 0 . (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 (0) = y 0 (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 φ ( 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 . 0-3
Thus, for the general solution, one needs to derive another type of the second solution . For this purpose, one may use the method of reduction of order . Let us look for a solution of the form C ( x ) e r 1 x with the undetermined function C ( x ) . By substituting the equation, we derive that L C ( x ) e r 1 x · = C ( x ) φ ( r 1 ) e r 1 x + a h C 00 ( x ) + 2 r 1 C 0 ( x ) i e r 1 x + bC 0 ( x ) e r 1 x = 0 .

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