Second-Order Linear Differential Equations A second-order linear differential equation has the form P͑x͒ d 2y dy ϩ R͑x͒y G͑x͒ ϩ Q͑x͒ dx dx where P, Q, R, and G are continuous functions We saw in Section 7.1 that 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 Such equations are called homogeneous linear equations Thus, the form of a second-order linear homogeneous differential equation is P͑x͒ d 2y dy ϩ R͑x͒y ϩ Q͑x͒ dx dx If G͑x͒ for some x, Equation 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 c1 y1 ϩ c2 y2 is also a solution Theorem If y1͑x͒ and y2͑x͒ are both solutions of the linear homogeneous equation (2) and c1 and c2 are any constants, then the function y͑x͒ c1 y1͑x͒ ϩ c2 y2͑x͒ is also a solution of Equation Proof Since y1 and y2 are solutions of Equation 2, we have P͑x͒y1Љ ϩ Q͑x͒y1Ј ϩ R͑x͒y1 and P͑x͒y2Љ ϩ Q͑x͒y2Ј ϩ R͑x͒y2 Therefore, using the basic rules for differentiation, we have P͑x͒yЉ ϩ Q͑x͒yЈ ϩ R͑x͒y P͑x͒͑c1 y1 ϩ c2 y2͒Љ ϩ Q͑x͒͑c1 y1 ϩ c2 y2͒Ј ϩ R͑x͒͑c1 y1 ϩ c2 y2͒ P͑x͒͑c1 y1Љ ϩ c2 y2Љ͒ ϩ Q͑x͒͑c1 y1Ј ϩ c2 y2Ј͒ ϩ R͑x͒͑c1 y1 ϩ c2 y2͒ c1͓P͑x͒y1Љ ϩ Q͑x͒yЈ1 ϩ R͑x͒y1͔ ϩ c2 ͓P͑x͒y2Љ ϩ Q͑x͒yЈ2 ϩ R͑x͒y2͔ c1͑0͒ ϩ c2͑0͒ Thus, y c1 y1 ϩ c2 y2 is a solution of Equation 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 and t͑x͒ 5x are linearly dependent, but f ͑x͒ e x and t͑x͒ xe x are linearly independent ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 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 c1 y1͑x͒ ϩ c2 y2͑x͒ where c1 and c2 are arbitrary constants Theorem is very useful because it says that if we know two particular linearly independent solutions, then we know every solution In general, it is not easy to discover particular solutions to a second-order linear equation But it is always possible to so if the coefficient functions P, Q, and R are constant functions, that is, if the differential equation has the form ayЉ ϩ byЈ ϩ cy where a, b, and c are constants and a It’s not hard to think of some likely candidates for particular solutions of Equation 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 We know that the exponential function y e rx (where r is a constant) has the property 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 ͑ar ϩ br ϩ c͒e rx or But e rx is never Thus, y e rx is a solution of Equation if r is a root of the equation ar ϩ br ϩ c Equation is called the auxiliary equation (or characteristic equation) of the differential equation ayЉ ϩ byЈ ϩ cy 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 Sometimes the roots r1 and r of the auxiliary equation can be found by factoring In other cases they are found by using the quadratic formula: r1 Ϫb ϩ sb Ϫ 4ac 2a r2 Ϫb Ϫ sb Ϫ 4ac 2a We distinguish three cases according to the sign of the discriminant b Ϫ 4ac b2 Ϫ 4ac Ͼ In this case the roots r1 and r of the auxiliary equation are real and distinct, so y1 e r x and y2 e r x are two linearly independent solutions of Equation (Note that e r x is not a constant multiple of e r x.) Therefore, by Theorem 4, we have the following fact CASE I ■ If the roots r1 and r of the auxiliary equation ar ϩ br ϩ c are real and unequal, then the general solution of ayЉ ϩ byЈ ϩ cy is y c1 e r x ϩ c2 e r x SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ In Figure the graphs of the basic solutions f ͑x͒ e x and t͑x͒ eϪ3 x of the differential equation in Example are shown in black and red, respectively Some of the other solutions, linear combinations of f and t , are shown in blue ■ ■ EXAMPLE Solve the equation yЉ ϩ yЈ Ϫ 6y SOLUTION The auxiliary equation is r ϩ r Ϫ ͑r Ϫ 2͒͑r ϩ 3͒ whose roots are r 2, Ϫ3 Therefore, by (8) the general solution of the given differential equation is 5f+g y c1 e 2x ϩ c2 eϪ3x f+5g f+g f g _1 g-f f-g _5 We could verify that this is indeed a solution by differentiating and substituting into the differential equation EXAMPLE Solve FIGURE d 2y dy Ϫ y ϩ dx dx SOLUTION To solve the auxiliary equation 3r ϩ r Ϫ we use the quadratic formula: r Ϫ1 Ϯ s13 Since the roots are real and distinct, the general solution is y c1 e (Ϫ1ϩs13 ) x͞6 ϩ c2 e (Ϫ1Ϫs13 ) x͞6 b Ϫ ac 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 r Then, from Equations 7, we have CASE II ■ rϪ b 2a so 2ar ϩ b We know that y1 e rx is one solution of Equation We now verify that y2 xe rx is also a solution: ay2Љ ϩ by2Ј ϩ cy2 a͑2re rx ϩ r 2xe rx ͒ ϩ b͑e rx ϩ rxe rx ͒ ϩ cxe rx ͑2ar ϩ b͒e rx ϩ ͑ar ϩ br ϩ c͒xe rx 0͑e rx ͒ ϩ 0͑xe rx ͒ The first term is by Equations 9; the second term is because r is a root of the auxiliary equation Since y1 e rx and y2 xe rx are linearly independent solutions, Theorem provides us with the general solution If the auxiliary equation ar ϩ br ϩ c has only one real root r, then the general solution of ayЉ ϩ byЈ ϩ cy is 10 y c1 e rx ϩ c2 xe rx EXAMPLE Solve the equation 4yЉ ϩ 12yЈ ϩ 9y SOLUTION The auxiliary equation 4r ϩ 12r ϩ can be factored as ͑2r ϩ 3͒2 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS so the only root is r Ϫ 32 By (10) the general solution is Figure shows the basic solutions f ͑x͒ eϪ3x͞2 and t͑x͒ xeϪ3x͞2 in Example and some other members of the family of solutions Notice that all of them approach as x l ϱ ■ ■ y c1 eϪ3x͞2 ϩ c2 xeϪ3x͞2 b Ϫ 4ac Ͻ In this case the roots r1 and r of the auxiliary equation are complex numbers (See Appendix I for information about complex numbers.) We can write CASE III f-g f ■ 5f+g f+5g _2 r1 ␣ ϩ i f+g g-f g r ␣ Ϫ i where ␣ and  are real numbers [In fact, ␣ Ϫb͑͞2a͒,  s4ac Ϫ b 2͑͞2a͒.] Then, using Euler’s equation _5 e i cos ϩ i sin FIGURE from Appendix I, we write the solution of the differential equation as y C1 e r x ϩ C2 e r x C1 e ͑␣ϩi͒x ϩ C2 e ͑␣Ϫi͒x C1 e ␣ x͑cos  x ϩ i sin  x͒ ϩ C2 e ␣ x͑cos  x Ϫ i sin  x͒ e ␣ x ͓͑C1 ϩ C2 ͒ cos  x ϩ i͑C1 Ϫ C2 ͒ sin  x͔ e ␣ x͑c1 cos  x ϩ c2 sin  x͒ where c1 C1 ϩ C2 , c2 i͑C1 Ϫ C2͒ This gives all solutions (real or complex) of the differential equation The solutions are real when the constants c1 and c2 are real We summarize the discussion as follows If the roots of the auxiliary equation ar ϩ br ϩ c are the complex numbers r1 ␣ ϩ i, r ␣ Ϫ i, then the general solution of ayЉ ϩ byЈ ϩ cy is 11 y e ␣ x͑c1 cos  x ϩ c2 sin  x͒ ■ ■ Figure shows the graphs of the solutions in Example 4, f ͑x͒ e x cos 2x and t͑x͒ e x sin 2x, together with some linear combinations All solutions approach as x l Ϫϱ f+g g f-g EXAMPLE Solve the equation yЉ Ϫ 6yЈ ϩ 13y SOLUTION The auxiliary equation is r Ϫ 6r ϩ 13 By the quadratic formula, the roots are r Ϯ s36 Ϫ 52 Ϯ sϪ16 Ϯ 2i 2 By (11) the general solution of the differential equation is f _3 y e 3x͑c1 cos 2x ϩ c2 sin 2x͒ Initial-Value and Boundary-Value Problems _3 FIGURE An initial-value problem for the second-order Equation or consists of finding a solution y of the differential equation that also satisfies initial conditions of the form y͑x ͒ y0 yЈ͑x ͒ y1 where y0 and y1 are given constants If P, Q, R, and G are continuous on an interval and P͑x͒ there, then a theorem found in more advanced books guarantees the existence and uniqueness of a solution to this initial-value problem Examples and illustrate the technique for solving such a problem SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ EXAMPLE Solve the initial-value problem yЉ ϩ yЈ Ϫ 6y ■ ■ Figure shows the graph of the solution of the initial-value problem in Example Compare with Figure y͑0͒ yЈ͑0͒ SOLUTION From Example we know that the general solution of the differential equa- tion is y͑x͒ c1 e 2x ϩ c2 eϪ3x 20 Differentiating this solution, we get yЈ͑x͒ 2c1 e 2x Ϫ 3c2 eϪ3x To satisfy the initial conditions we require that _2 FIGURE 12 y͑0͒ c1 ϩ c2 13 yЈ͑0͒ 2c1 Ϫ 3c2 From (13) we have c2 23 c1 and so (12) gives c1 ϩ 23 c1 c1 35 c2 25 Thus, the required solution of the initial-value problem is y 35 e 2x ϩ 25 eϪ3x ■ ■ The solution to Example is graphed in Figure It appears to be a shifted sine curve and, indeed, you can verify that another way of writing the solution is y s13 sin͑x ϩ ͒ where tan 23 EXAMPLE Solve the initial-value problem yЉ ϩ y y͑0͒ yЈ͑0͒ SOLUTION The auxiliary equation is r ϩ 0, or r Ϫ1, whose roots are Ϯi Thus ␣ 0,  1, and since e 0x 1, the general solution is y͑x͒ c1 cos x ϩ c2 sin x _2π 2π yЈ͑x͒ Ϫc1 sin x ϩ c2 cos x Since the initial conditions become _5 FIGURE y͑0͒ c1 yЈ͑0͒ c2 Therefore, the solution of the initial-value problem is y͑x͒ cos x ϩ sin x A boundary-value problem for Equation consists of finding a solution y of the differential equation that also satisfies boundary conditions of the form y͑x ͒ y0 y͑x ͒ y1 In contrast with the situation for initial-value problems, a boundary-value problem does not always have a solution EXAMPLE Solve the boundary-value problem yЉ ϩ 2yЈ ϩ y y͑0͒ y͑1͒ SOLUTION The auxiliary equation is r ϩ 2r ϩ or ͑r ϩ 1͒2 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS whose only root is r Ϫ1 Therefore, the general solution is ■ ■ Figure shows the graph of the solution of the boundary-value problem in Example y͑x͒ c1 eϪx ϩ c2 xeϪx The boundary conditions are satisfied if _1 y͑0͒ c1 y͑1͒ c1 eϪ1 ϩ c2 eϪ1 The first condition gives c1 1, so the second condition becomes _5 eϪ1 ϩ c2 eϪ1 FIGURE Solving this equation for c2 by first multiplying through by e, we get ϩ c2 3e c2 3e Ϫ so Thus, the solution of the boundary-value problem is y eϪx ϩ ͑3e Ϫ 1͒xeϪx Summary: Solutions of ayЈЈ ϩ byЈ ϩ c ϭ Roots of ar ϩ br ϩ c General solution y c1 e r x ϩ c2 e r x y c1 e rx ϩ c2 xe rx y e ␣ x͑c1 cos  x ϩ c2 sin  x͒ r1, r2 real and distinct r1 r2 r r1, r2 complex: ␣ Ϯ i Exercises A Click here for answers 1–13 S 17–24 Click here for solutions 17 2yЉ ϩ 5yЈ ϩ 3y 0, 18 yЉ ϩ 3y 0, Solve the differential equation y͑0͒ 3, y͑0͒ 1, yЈ͑0͒ Ϫ4 yЈ͑0͒ yЉ Ϫ 6yЈ ϩ 8y yЉ Ϫ 4yЈ ϩ 8y 19 yЉ Ϫ yЈ ϩ y 0, yЉ ϩ 8yЈ ϩ 41y 2yЉ Ϫ yЈ Ϫ y 20 2yЉ ϩ 5yЈ Ϫ 3y 0, yЉ Ϫ 2yЈ ϩ y 3yЉ 5yЈ 21 yЉ ϩ 16y 0, 4yЉ ϩ y 16yЉ ϩ 24yЈ ϩ 9y 22 yЉ Ϫ 2yЈ ϩ 5y 0, y͑͒ 0, yЈ͑͒ 10 9yЉ ϩ 4y 23 yЉ ϩ 2yЈ ϩ 2y 0, y͑0͒ 2, yЈ͑0͒ d 2y dy Ϫ6 ϩ 4y 12 dt dt 24 yЉ ϩ 12yЈ ϩ 36y 0, 4yЉ ϩ yЈ d y dy Ϫ2 Ϫy0 11 dt dt ■ d 2y dy ϩ ϩy0 13 dt dt ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 25 yЉ ϩ y 0, ■ ■ ■ ■ ■ ■ yЈ͑0͒ yЈ͑͞4͒ y͑1͒ 0, ■ y͑0͒ 1, ■ yЈ͑1͒ ■ ■ ■ 31 yЉ ϩ 4yЈ ϩ 13y 0, y͑͒ Ϫ4 y͑1͒ y͑0͒ 1, y͑0͒ 2, 30 yЉ Ϫ 6yЈ ϩ 9y 0, ■ y͑0͒ 1, y͑0͒ 3, 29 yЉ Ϫ 6yЈ ϩ 25y 0, d 2y dy Ϫ2 ϩ 5y 16 dx dx ■ ■ yЈ͑0͒ Ϫ1.5 ■ Solve the boundary-value problem, if possible 28 yЉ ϩ 100 y 0, d y dy Ϫ8 ϩ 16y dx dx 15 ■ y͑0͒ 1, y͑͞4͒ Ϫ3, 27 yЉ Ϫ 3yЈ ϩ 2y 0, d y dy Ϫ Ϫ 2y dx dx ■ 26 yЉ ϩ 2yЈ 0, Graph the two basic solutions of the differential equation and several other solutions What features the solutions have in common? 14 ■ 25–32 ; 14–16 ■ Solve the initial-value problem y͑3͒ y͑͒ y͑0͒ 1, y͑0͒ 1, y͑0͒ 2, y͑͒ y͑1͒ y͑ ͞2͒ ■ ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ 32 9yЉ Ϫ 18yЈ ϩ 10 y 0, ■ ■ ■ ■ ■ ■ y͑0͒ 0, ■ ■ y͑͒ ■ ■ ■ ■ 33 Let L be a nonzero real number (a) Show that the boundary-value problem yЉ ϩ y 0, y͑0͒ 0, y͑L͒ has only the trivial solution y for the cases and Ͻ ■ (b) For the case Ͼ 0, find the values of for which this problem has a nontrivial solution and give the corresponding solution 34 If a, b, and c are all positive constants and y͑x͒ is a solution of the differential equation ayЉ ϩ byЈ ϩ cy 0, show that lim x l ϱ y͑x͒ 8 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS Answers S Click here for solutions y c1 e 4x ϩ c2 e 2x y eϪ4x ͑c1 cos 5x ϩ c2 sin 5x͒ x x y c1 e ϩ c2 xe y c1 cos͑x͞2͒ ϩ c2 sin͑x͞2͒ y c1 ϩ c2 eϪx͞4 11 y c1 e (1ϩs2 )t ϩ c2 e (1Ϫs2 )t 13 y eϪt͞2 [c1 cos(s3t͞2) ϩ c2 sin(s3t͞2)] g 15 40 f _0.2 _40 All solutions approach as x l Ϫϱ and approach Ϯϱ as x l ϱ 17 y 2eϪ3x͞2 ϩ eϪx 19 y e x/2 Ϫ 2xe x͞2 21 y cos 4x Ϫ sin 4x 23 y eϪx͑2 cos x ϩ sin x͒ e 2x e xϩ3 1 25 y cos( x) Ϫ sin( x) 27 y ϩ e Ϫ1 Ϫ e3 29 No solution 31 y eϪ2x ͑2 cos 3x Ϫ e sin 3x͒ 33 (b) n 2 2͞L2, n a positive integer; y C sin͑n x͞L͒ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS ■ Solutions: Second-Order Linear Differential Equations The auxiliary equation is r − 6r + = ⇒ solution is y = c1 e4x + c2 e2x (r − 4)(r − 2) = ⇒ r = 4, r = Then by (8) the general The auxiliary equation is r2 + 8r + 41 = ⇒ r = −4 ± 5i Then by (11) the general solution is y = e−4x (c1 cos 5x + c2 sin 5x) The auxiliary equation is r2 − 2r + = (r − 1)2 = y = c1 ex + c2 xex The auxiliary equation is 4r + = ⇒ r = Then by (10), the general solution is r = ± 12 i, so y = c1 cos ⇒ The auxiliary equation is 4r + r = r(4r + 1) = ⇒ ¡1 ¢ ¡1 ¢ x + c2 sin x r = 0, r = − 14 , so y = c1 + c2 e−x/4 √ √ √ 2, so y = c1 e(1+ 2)t + c2 e(1− 2)t ³√ ´ ³ √ ´i h √ ⇒ r = − 12 ± 23 i, so y = e−t/2 c1 cos 23 t + c2 sin 23 t 11 The auxiliary equation is r − 2r − = ⇒ r = ± 13 The auxiliary equation is r + r + = 15 r − 8r + 16 = (r − 4)2 = so y = c1 e4x + c2 xe4x The graphs are all asymptotic to the x-axis as x → −∞, and as x → ∞ the solutions tend to ±∞ 17 2r + 5r + = (2r + 3)(r + 1) = 0, so r = − 32 , r = −1 and the general solution is y = c1 e−3x/2 + c2 e−x Then y(0) = ⇒ c1 + c2 = and y0 (0) = −4 ⇒ − 32 c1 − c2 = −4, so c1 = and c2 = Thus the solution to the initial-value problem is y = 2e−3x/2 + e−x 19 4r − 4r + = (2r − 1)2 = ⇒ r = ⇒ c1 = and y0 (0) = −1.5 ⇒ y=e x/2 x/2 − 2xe c1 and the general solution is y = c1 ex/2 + c2 xex/2 Then y(0) = + c2 = −1.5, so c2 = −2 and the solution to the initial-value problem is 21 r + 16 = ⇒ r = ±4i and the general solution is y = e0x (c1 cos 4x + c2 sin 4x) = c1 cos 4x + c2 sin 4x ¡ ¢ ¡ ¢ Then y π4 = −3 ⇒ −c1 = −3 ⇒ c1 = and y π4 = ⇒ −4c2 = ⇒ c2 = −1, so the solution to the initial-value problem is y = cos 4x − sin 4x 23 r2 + 2r + = ⇒ r = −1 ± i and the general solution is y = e−x (c1 cos x + c2 sin x) Then = y(0) = c1 c2 = and the solution to the initial-value problem is y = e−x (2 cos x + sin x) ¡ ¢ ¡ ¢ 25 4r2 + = ⇒ r = ± 12 i and the general solution is y = c1 cos 12 x + c2 sin 12 x Then = y(0) = c1 and ¡ ¢ ¡ ¢ −4 = y(π) = c2 , so the solution of the boundary-value problem is y = cos 12 x − sin 12 x and = y (0) = c2 − c1 ⇒ 27 r − 3r + = (r − 2)(r − 1) = ⇒ r = 1, r = and the general solution is y = c1 ex + c2 e2x Then = y(0) = c1 + c2 and = y(3) = c1 e3 + c2 e6 so c2 = 1/(1 − e3 ) and c1 = e3 /(e3 − 1) The solution of the boundary-value problem is y = 29 r − 6r + 25 = ⇒ and = y(π) = c1 e3π e2x ex+3 + e3 − 1 − e3 r = ± 4i and the general solution is y = e3x (c1 cos 4x + c2 sin 4x) But = y(0) = c1 ⇒ c1 = 2/e3π , so there is no solution 10 ■ SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS r = −2 ± 3i and the general solution is y = e−2x (c1 cos 3x + c2 sin 3x) But ¡ ¢ = y(0) = c1 and = y π2 = e−π (−c2 ), so the solution to the boundary-value problem is 31 r + 4r + 13 = ⇒ y = e−2x (2 cos 3x − eπ sin 3x) 33 (a) Case (λ = 0): y 00 + λy = ⇒ y 00 = which has an auxiliary equation r = ⇒ r = ⇒ y = c1 + c2 x where y(0) = and y(L) = Thus, = y(0) = c1 and = y(L) = c2 L ⇒ c1 = c2 = Thus, y = √ Case (λ < 0): y 00 + λy = has auxiliary equation r2 = −λ ⇒ r = ± −λ (distinct and real since λ < 0) ⇒ y = c1 e √ = y(L) = c1 e −λL Multiplying (∗) by e √ √ √ −λx + c e− √ − −λL + c2 e −λL −λx where y(0) = and y(L) = Thus, = y(0) = c1 + c2 (∗) and (†) ³ √ ´ √ and subtracting (†) gives c2 e −λL − e− −λL = ⇒ c2 = and thus c1 = from (∗) Thus, y = for the cases λ = and λ < √ √ ⇒ y = c1 cos λ x + c2 sin λ x √ where y(0) = and y(L) = Thus, = y(0) = c1 and = y(L) = c2 sin λL since c1 = Since we √ √ λ L = nπ where n is an integer cannot have a trivial solution, c2 6= and thus sin λ L = ⇒ (b) y 00 + λy = has an auxiliary equation r + λ = ⇒ ⇒ r = ±i √ λ λ = n2 π /L2 and y = c2 sin(nπx/L) where n is an integer