SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS A second-order linear differential equation has the form d 2y dy 1 P͑x͒ ϩ Q͑x͒ ϩ R͑x͒y ෇ G͑x͒ dx2 dx where P,Q ,R , and G are continuous functions. Equations of this type arise in the study of the motion of a spring. In Additional Topics: Applications of Second-Order Differential Equations we will further pursue this application as well as the application to electric circuits. In this section we study the case where G͑x͒ ෇ 0, for all x, in Equation 1. Such equa- tions are called homogeneous linear equations. Thus, the form of a second-order linear homogeneous differential equation is d 2 y dy 2 P͑x͒ ϩ Q͑x͒ ϩ R͑x͒y ෇ 0 dx2 dx If G͑x͒ 0 for some x, Equation 1 is nonhomogeneous and is discussed in Additional Topics: Nonhomogeneous Linear Equations. Two basic facts enable us to solve homogeneous linear equations. The first of these says that if we know two solutions y1 and y2 of such an equation, then the linear combination y ෇ c1y1 ϩ c2y2 is also a solution. 3 Theorem If y1͑x͒ and y2͑x͒ are both solutions of the linear homogeneous equa- tion (2) and c1 and c2 are any constants, then the function y͑x͒ ෇ c1y1͑x͒ ϩ c2y2͑x͒ is also a solution of Equation 2. Proof Since y1 and y2 are solutions of Equation 2, we have P͑x͒y1ЉϩQ͑x͒y1ЈϩR͑x͒y1 ෇ 0 and P͑x͒y2ЉϩQ͑x͒y2ЈϩR͑x͒y2 ෇ 0 Therefore, using the basic rules for differentiation, we have P͑x͒yЉϩQ͑x͒yЈϩR͑x͒y ෇ P͑x͒͑c1y1 ϩ c2y2͒ЉϩQ͑x͒͑c1y1 ϩ c2y2͒ЈϩR͑x͒͑c1y1 ϩ c2y2͒ ෇ P͑x͒͑c1y1Љϩc2y2Љ͒ ϩ Q͑x͒͑c1y1Јϩc2y2Ј͒ ϩ R͑x͒͑c1y1 ϩ c2y2͒ ෇ c1͓P͑x͒y1ЉϩQ͑x͒y1ЈϩR͑x͒y1͔ ϩ c2 ͓P͑x͒y2ЉϩQ͑x͒y2ЈϩR͑x͒y2͔ ෇ c1͑0͒ ϩ c2͑0͒ ෇ 0 Thus,y ෇ c1y1 ϩ c2y2 is a solution of Equation 2. The other fact we need is given by the following theorem, which is proved in more advanced courses. It says that the general solution is a linear combination of two linearly independent solutions y1 and y2. This means that neither y1 nor y2 is a constant multiple of the other. For instance, the functions f ͑x͒ ෇ x 2 and t͑x͒ ෇ 5x 2 are linearly dependent, but f ͑x͒ ෇ e x and t͑x͒ ෇ xe x are linearly independent. 1 Thomson Brooks-Cole copyright 2007 2 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 4 Theorem If y1 and y2 are linearly independent solutions of Equation 2, and P͑x͒ is never 0, then the general solution is given by y͑x͒ ෇ c1y1͑x͒ ϩ c2y2͑x͒ where c1 and c2 are arbitrary constants. Theorem 4 is very useful because it says that if we know two particular linearly inde- pendent solutions, then we know every solution. In general, it is not easy to discover particular solutions to a second-order linear equa- tion. But it is always possible to do so if the coefficient functions P,Q , and R are constant functions, that is, if the differential equation has the form 5 ayЉϩbyЈϩcy ෇ 0 where a,b , and c are constants and a 0. It’s not hard to think of some likely candidates for particular solutions of Equation 5 if we state the equation verbally. We are looking for a function y such that a constant times its second derivative yЉ plus another constant times yЈ plus a third constant times y is equal to 0. We know that the exponential function y ෇ e rx (where r is a constant) has the prop- erty that its derivative is a constant multiple of itself:yЈ ෇ re rx . Furthermore,yЉ ෇ r 2e rx . If we substitute these expressions into Equation 5, we see that y ෇ e rx is a solution if ar 2e rx ϩ bre rx ϩ ce rx ෇ 0 or ͑ar 2 ϩ br ϩ c͒e rx ෇ 0 But e rx is never 0. Thus,y ෇ e rx is a solution of Equation 5 if r is a root of the equation 6 ar 2 ϩ br ϩ c ෇ 0 Equation 6 is called the auxiliary equation (or characteristic equation) of the differen- tial equation ayЉϩbyЈϩcy ෇ 0. Notice that it is an algebraic equation that is obtained from the differential equation by replacing yЉ by r 2,yЈ by r, and y by 1. Sometimes the roots r1 and r2 of the auxiliary equation can be found by factoring. In other cases they are found by using the quadratic formula: Ϫb ϩ sb 2 Ϫ 4ac Ϫb Ϫ sb 2 Ϫ 4ac 7 r1 ෇ r2 ෇ 2a 2a We distinguish three cases according to the sign of the discriminant b 2 Ϫ 4ac. 2 CASE I ■ b Ϫ 4ac Ͼ 0 r1x In this case the roots r1 and r2 of the auxiliary equation are real and distinct, so y1 ෇ e r2 x r2 x and y2 ෇ e are two linearly independent solutions of Equation 5. (Note that e is not a constant multiple of e r1x.) Therefore, by Theorem 4, we have the following fact. 2 8 If the roots r1 and r2 of the auxiliary equation ar ϩ br ϩ c ෇ 0 are real and unequal, then the general solution of ayЉϩbyЈϩcy ෇ 0 is r1x r2 x y ෇ c1e ϩ c2 e Thomson Brooks-Cole copyright 2007 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 3 ■■ In Figure 1 the graphs of the basic solutions EXAMPLE 1 Solve the equation yЉϩyЈϪ6y ෇ 0. f ͑x͒ ෇ e 2x and t͑x͒ ෇ eϪ3x of the differential equation in Example 1 are shown in black and SOLUTION The auxiliary equation is red, respectively. Some of the other solutions, 2 linear combinations of f and t, are shown r ϩ r Ϫ 6 ෇ ͑r Ϫ 2͒͑r ϩ 3͒ ෇ 0 in blue. whose roots are r ෇ 2,Ϫ3 . Therefore, by (8) the general solution of the given differen- 8 tial equation is 5f+g f+5g 2x Ϫ3x y ෇ c1e ϩ c2 e f+g fgWe could verify that this is indeed a solution by differentiating and substituting into the _1 1 g-f differential equation. f-g 2 _5 d y ϩ dy Ϫ ෇ EXAMPLE 2 Solve 3 2 y 0. FIGURE 1 dx dx SOLUTION To solve the auxiliary equation 3r 2 ϩ r Ϫ 1 ෇ 0 we use the quadratic formula: Ϫ1 Ϯ s13 r ෇ 6 Since the roots are real and distinct, the general solution is (Ϫ1ϩs13 )x͞6 (Ϫ1Ϫs13 )x͞6 y ෇ c1e ϩ c2 e 2 CASE II ■ b Ϫ 4ac ෇ 0 In this case r1 ෇ r2; that is, the roots of the auxiliary equation are real and equal. Let’s denote by r the common value of r1 and r2. Then, from Equations 7, we have b 9 r ෇ Ϫ so 2ar ϩ b ෇ 0 2a rx rx We know that y1 ෇ e is one solution of Equation 5. We now verify that y2 ෇ xe is also a solution: rx 2 rx rx rx rx ay2Љϩby2Јϩcy2 ෇ a͑2re ϩ r xe ͒ ϩ b͑e ϩ rxe ͒ ϩ cxe ෇ ͑2ar ϩ b͒e rx ϩ ͑ar 2 ϩ br ϩ c͒xe rx ෇ 0͑e rx͒ ϩ 0͑xe rx͒ ෇ 0 The first term is 0 by Equations 9; the second term is 0 because r is a root of the auxiliary rx rx equation. Since y1 ෇ e and y2 ෇ xe are linearly independent solutions, Theorem 4 pro- vides us with the general solution. 10 If the auxiliary equation ar 2 ϩ br ϩ c ෇ 0 has only one real root r, then the general solution of ayЉϩbyЈϩcy ෇ 0 is rx rx y ෇ c1e ϩ c2 xe EXAMPLE 3 Solve the equation 4yЉϩ12yЈϩ9y ෇ 0. SOLUTION The auxiliary equation 4r 2 ϩ 12r ϩ 9 ෇ 0 can be factored as ͑2r ϩ 3͒2 ෇ 0 Thomson Brooks-Cole copyright 2007 4 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■■ Figure 2 shows the basic solutions ෇ Ϫ 3 so the only root is r 2 . By (10) the general solution is f ͑x͒ ෇ eϪ3x͞2 and t͑x͒ ෇ xeϪ3x͞2 in Example 3 and some other members of the y ෇ c eϪ3x͞2 ϩ c xeϪ3x͞2 family of solutions. Notice that all of them 1 2 approach 0 as x l ϱ. 2 CASE III ■ b Ϫ 4ac Ͻ 0 f-g 8 In this case the roots r1 and r2 of the auxiliary equation are complex numbers. (See f Additional Topics: Complex Numbers for information about complex numbers.) We can 5f+g write f+5g r1 ෇ ␣ ϩ i␤ r2 ෇ ␣ Ϫ i␤ _2 2 f+g g-f g where ␣ and ␤ are real numbers. [In fact,␣ ෇ Ϫb͑͞2a͒ ,␤ ෇ s4ac Ϫ b 2͑͞2a͒ .] Then, using Euler’s equation _5 i␪ FIGURE 2 e ෇ cos ␪ ϩ i sin ␪ from Additional Topics: Complex Numbers, we write the solution of the differential equa- tion as r1x r2 x ͑␣ϩi␤͒x ͑␣Ϫi␤͒x y ෇ C1e ϩ C2e ෇ C1e ϩ C2e ␣ x ␣ x ෇ C1e ͑cos ␤x ϩ i sin ␤x͒ ϩ C2e ͑cos ␤x Ϫ i sin ␤x͒ ␣ x ෇ e ͓͑C1 ϩ C2 ͒ cos ␤x ϩ i͑C1 Ϫ C2 ͒ sin ␤x͔ ␣ x ෇ e ͑c1 cos ␤x ϩ c2 sin ␤x͒ where c1 ෇ C1 ϩ C2,c2 ෇ i͑C1 Ϫ C2͒ . This gives all solutions (real or complex) of the dif- ferential equation. The solutions are real when the constants c1 and c2 are real. We sum- marize the discussion as follows. 11 If the roots of the auxiliary equation ar 2 ϩ br ϩ c ෇ 0 are the complex num- bers r1 ෇ ␣ ϩ i␤,r2 ෇ ␣ Ϫ i␤ , then the general solution of ayЉϩbyЈϩcy ෇ 0 is ␣ x y ෇ e ͑c1 cos ␤x ϩ c2 sin ␤x͒ ■■ Figure 3 shows the graphs of the solu- EXAMPLE 4 Solve the equation yЉϪ6yЈϩ13y ෇ 0. tions in Example 4, f ͑x͒ ෇ e 3x cos 2x and 2 t͑x͒ ෇ e 3x sin 2x, together with some linear SOLUTION The auxiliary equation is r Ϫ 6r ϩ 13 ෇ 0.