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# ch5 - CHAPTER 5 HIGHER-ORDER LINEAR DIFFERENTIAL EQUATIONS...

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232 Chapter 5 CHAPTER 5 HIGHER-ORDER LINEAR DIFFERENTIAL EQUATIONS SECTION 5.1 INTRODUCTION: SECOND-ORDER LINEAR EQUATIONS In this section the central ideas of the theory of linear differential equations are introduced and illustrated concretely in the context of second-order equations. These key concepts include superposition of solutions (Theorem 1), existence and uniqueness of solutions (Theorem 2), linear independence, the Wronskian (Theorem 3), and general solutions (Theorem 4). This discussion of second-order equations serves as preparation for the treatment of n th order linear equations in Section 5.2. Although the concepts in this section may seem somewhat abstract to students, the problems set is quite tangible and largely computational. In each of Problems 1–16 the verification that y 1 and y 2 satisfy the given differential equation is a routine matter. As in E x ample 2, we then impose the given initial conditions on the general solution y = c 1 y 1 + c 2 y 2 . This yields two linear equations that determine the values of the constants c 1 and c 2 . 1. Imposition of the initial conditions (0) 0, (0) 5 y y = = on the general solution 1 2 ( ) x x y x c e c e = + yields the two equations 1 2 1 2 0, 0 c c c c + = = with solution 1 2 5/ 2, 5/ 2. c c = = − Hence the desired particular solution is y ( x ) = 5( e x - e - x )/2. 2. Imposition of the initial conditions (0) 1, (0) 15 y y = − = on the general solution 3 3 1 2 ( ) x x y x c e c e = + yields the two equations 1 2 1 2 1, 3 3 15 c c c c + = − = with solution 1 2 2, 3. c c = = Hence the desired particular solution is y ( x ) = 2 e 3 x - 3 e - 3 x . 3. Imposition of the initial conditions (0) 3, (0) 8 y y = = on the general solution 1 2 ( ) cos2 sin2 y x c x c x = + yields the two equations 1 2 3, 2 8 c c = = with solution 1 2 3, 4. c c = = Hence the desired particular solution is y ( x ) = 3 cos 2 x + 4 sin 2 x . 4. Imposition of the initial conditions (0) 10, (0) 10 y y = = − on the general solution 1 2 ( ) cos5 sin5 y x c x c x = + yields the two equations 1 2 10, 5 10 c c = = − with solution 1 2 3, 4. c c = = Hence the desired particular solution is y ( x ) = 10 cos 5 x - 2 sin 5 x.

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Section 5.1 233 5. Imposition of the initial conditions (0) 1, (0) 0 y y = = on the general solution 2 1 2 ( ) x x y x c e c e = + yields the two equations 1 2 1 2 1, 2 0 c c c c + = + = with solution 1 2 2, 1. c c = = − Hence the desired particular solution is y ( x ) = 2 e x - e 2 x . 6. Imposition of the initial conditions (0) 7, (0) 1 y y = = − on the general solution 2 3 1 2 ( ) x x y x c e c e = + yields the two equations 1 2 1 2 7, 2 3 1 c c c c + = = − with solution 1 2 4, 3. c c = = Hence the desired particular solution is y ( x ) = 4 e 2 x + 3 e - 3 x . 7. Imposition of the initial conditions (0) 2, (0) 8 y y = − = on the general solution 1 2 ( ) x y x c c e = + yields the two equations 1 2 2 2, 8 c c c + = − = with solution 1 2 6, 8. c c = = − Hence the desired particular solution is y ( x ) = 6 8 e - x .
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