SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS A second-order linear differential equation has the form 1 Px d y dx Qx dx Rxy Gx where P, Q, R, and G are continuous functions. Equations of this type arise in the stu 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 stu the case where Gx 0, for all x, in Equation 1. Such equations are called homogeneous linear equations. Thus, the form of a second-order linear homogeneous differential equation is Px d y dx Qx dx Rxy 0 If Gx 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 y 1 and y of such an equation, then the linear combination y c 1 y 1 c y is also a solution. 3 Theorem If y 1 x and y x are both solutions of the linear homogeneous equation () and and are any constants, then the function c 1 c yx c 1 y 1 x c y x is also a solution of Equation. y 1 y Proof Since and are solutions of Equation, we have Pxy 1 Qxy 1 Rxy 1 0 and Pxy Qxy Rxy 0 Therefore, using the basic rules for differentiation, we have Pxy Qxy Rxy Pxc 1 y 1 c y Qxc 1 y 1 c y Rxc 1 y 1 c y Pxc 1 y 1 c y Qxc 1 y 1 c y Rxc 1 y 1 c y c 1 Pxy 1 Qxy 1 Rxy 1 c Pxy Qxy Rxy c 1 0 c 0 0 Thus, y c 1 y 1 c y 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 y 1 and y. This means that neither y 1 nor y is a constant multiple of the other. For instance, the functions f x x and tx 5x are linearly dependent, but f x e x and tx xe x are linearly independent. 1

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 4 Theorem If y 1 and y are linearly independent solutions of Equation, and Px is never 0, then the general solution is given by yx c 1 y 1 x c y x c 1 c where and are arbitrary constants. Theorem 4 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 do so if the coefficient functions P, Q, and R are constant functions, that is, if the differential equation has the form 5 ay by cy 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 y plus another constant times y 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 property that its derivative is a constant multiple of itself: y re rx. Furthermore, y r e rx. If we substitute these expressions into Equation 5, we see that y e rx is a solution if ar e rx bre rx ce rx 0 or ar br ce 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 br c 0 Equation 6 is called the auxiliary equation (or characteristic equation) of the differential equation ay by cy 0. Notice that it is an algebraic equation that is obtained from the differential equation by replacing y by r, y by r, and y by 1. Sometimes the roots r 1 and r of the auxiliary equation can be found by factoring. In other cases they are found by using the quadratic formula: 7 r 1 b sb 4ac a r b sb 4ac a We distinguish three cases according to the sign of the discriminant b 4ac. CASE I b 4ac 0 In this case the roots r and of the auxiliary equation are real and distinct, so y 1 e r1x 1 r and are two linearly independent solutions of Equation 5. (Note that e r x y e r x is not a constant multiple of e r1x.) Therefore, by Theorem 4, we have the following fact. 8 If the roots r and of the auxiliary equation ar 1 r br c 0 are real and unequal, then the general solution of ay by cy 0 is y c 1 e r1x c e r x

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 3 In Figure 1 the graphs of the basic solutions f x e x and tx e 3x of the differential equation in Example 1 are shown in black and red, respectively. Some of the other solutions, linear combinations of f and t, are shown in blue. 8 5f+g f+5g f+g f g _1 1 g-f f-g EXAMPLE 1 Solve the equation y y 6y 0. SOLUTION The auxiliary equation is r r 6 r r 3 0 whose roots are r, 3. Therefore, by (8) the general solution of the given differential equation is y c 1 e x c e 3x We could verify that this is indeed a solution by differentiating and substituting into the differential equation. FIGURE 1 _5 EXAMPLE Solve 3 d y y 0. dx dx SOLUTION To solve the auxiliary equation 3r r 1 0 we use the quadratic formula: r 1 s13 6 Since the roots are real and distinct, the general solution is y c 1 e (1s13 )x6 (1s13 )x6 c e CASE II b 4ac 0 In this case r 1 r ; that is, the roots of the auxiliary equation are real and equal. Let s denote by r the common value of and r. Then, from Equations 7, we have r 1 9 r b a so ar b 0 We know that y is one solution of Equation 5. We now verify that y xe rx 1 e rx is also a solution: ay by cy are rx r xe rx be rx rxe rx cxe rx ar be rx ar br cxe rx 0e rx 0xe rx 0 The first term is 0 by Equations 9; the second term is 0 because r is a root of the auxiliary equation. Since y and y xe rx 1 e rx are linearly independent solutions, Theorem 4 provides us with the general solution. 10 If the auxiliary equation ar br c 0 has only one real root r, then the general solution of ay by cy 0 is y c 1 e rx c xe rx EXAMPLE 3 Solve the equation 4y 1y 9y0. SOLUTION The auxiliary equation 4r 1r 9 0 can be factored as r 3 0

4 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS Figure shows the basic solutions f x e 3x and tx xe 3x in Example 3 and some other members of the family of solutions. Notice that all of them approach 0 as x l. f FIGURE f-g 8 _5 5f+g f+5g _ f+g g-f g so the only root is r 3. By (10) the general solution is CASE III b 4ac 0 In this case the roots r 1 and r of the auxiliary equation are complex numbers. (See Additional Topics: Complex Numbers for information about complex numbers.) We can write y c 1 e 3x c xe 3x r 1 i r i ba where and are real numbers. [In fact,,.] Then, using Euler s equation e i cos i sin s4ac b a from Additional Topics: Complex Numbers, we write the solution of the differential equation as y C 1 e r1x C e r x C 1 e ix C e ix C 1 e x cos x i sin x C e x cos x i sin x e x C 1 C cos x ic 1 C sin x e x c 1 cos x c sin x where c 1 C 1 C, c ic 1 C. This gives all solutions (real or complex) of the differential equation. The solutions are real when the constants c 1 and c are real. We summarize the discussion as follows. 11 If the roots of the auxiliary equation ar br c 0 are the complex numbers r 1 i, r i, then the general solution of ay by cy 0 is y e x c 1 cos x c sin x Figure 3 shows the graphs of the solutions in Example 4, f x e 3x cos x and tx e 3x sin x, together with some linear combinations. All solutions approach 0 as x l. f+g f-g f _3 3 g EXAMPLE 4 Solve the equation y 6y 13y 0. SOLUTION The auxiliary equation is r 6r 13 0. By the quadratic formula, the roots are r 6 s36 5 By (11) the general solution of the differential equation is 6 s16 y e 3x c 1 cos x c sin x 3 i FIGURE 3 _3 INITIAL-VALUE AND BOUNDARY-VALUE PROBLEMS An initial-value problem for the second-order Equation 1 or consists of finding a solution y of the differential equation that also satisfies initial conditions of the form yx 0 y 0 yx 0 y 1 where y 0 and y 1 are given constants. If P, Q, R, and G are continuous on an interval and Px 0 there, then a theorem found in more advanced books guarantees the existence and uniqueness of a solution to this initial-value problem. Examples 5 and 6 illustrate the technique for solving such a problem.

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 5 EXAMPLE 5 Solve the initial-value problem Figure 4 shows the graph of the solution of the initial-value problem in Example 5. Compare with Figure 1. 0 y y 6y 0 y0 1 y0 0 SOLUTION From Example 1 we know that the general solution of the differential equation is yx c 1 e x c e 3x Differentiating this solution, we get yx c 1 e x 3c e 3x To satisfy the initial conditions we require that _ 0 FIGURE 4 1 y0 c 1 c 1 13 y0 c 1 3c 0 From (13) we have c 3c 1 and so (1) gives c 1 3c 1 1 c 1 3 5 Thus, the required solution of the initial-value problem is c 5 y 3 5e x 5 e 3x The solution to Example 6 is graphed in Figure 5. It appears to be a shifted sine curve and, indeed, you can verify that another way of writing the solution is y s13 sinx where tan 3 _π 5 π EXAMPLE 6 Solve the initial-value problem y y 0 y0 y0 3 SOLUTION The auxiliary equation is r 1 0, or r 1, whose roots are i. Thus,, and since e 0x 1, the general solution is 0 1 yx c 1 cos x c sin x Since yx c 1 sin x c cos x the initial conditions become y0 c 1 y0 c 3 FIGURE 5 _5 Therefore, the solution of the initial-value problem is yx cos x 3sin x A boundary-value problem for Equation 1 consists of finding a solution y of the differential equation that also satisfies boundary conditions of the form yx 0 y 0 yx 1 y 1 In contrast with the situation for initial-value problems, a boundary-value problem does not always have a solution. EXAMPLE 7 Solve the boundary-value problem y y y 0 y0 1 y1 3 SOLUTION The auxiliary equation is r r 1 0 or r 1 0 whose only root is r 1. Therefore, the general solution is yx c 1 e x c xe x

6 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS Figure 6 shows the graph of the solution of the boundary-value problem in Example 7. 5 _1 5 _5 FIGURE 6 The boundary conditions are satisfied if y0 c 1 1 y1 c 1 e 1 c e 1 3 The first condition gives c 1 1, so the second condition becomes e1 c e1 3 Solving this equation for c by first multiplying through by e, we get 1 c 3e so c 3e 1 Thus, the solution of the boundary-value problem is Summary: Solutions of ay by c 0 y e x 3e 1xe x Roots of ar br c 0 r 1, r real and distinct r 1 r r r 1, r complex: i General solution y c 1 e r1x c e r x y c 1 e rx c xe rx y e x c 1 cos x c sin x EXERCISES 1 13 Solve the differential equation. 1. y 6y 8y0. y 4y 8y 0 3. y 8y 41y 0 4. y y y 0 5. y y y 0 6. 3y 5y 7. 4y y 0 8. 16y 4y 9y 0 9. 4y y 0 10. 9y 4y 0 11. d y d y y 0 1. 6 4y 0 dt dt dt dt 13. ; 14 16 Graph the two basic solutions of the differential equation and several other solutions. What features do the solutions have in common? 14. 6 d y y 0 15. dx dx 16. A d y dt Click here for answers. d y dx y 0 dt dx 5y 0 17 4 Solve the initial-value problem. 17. y 5y 3y 0, y0 3, 18. y 3y 0, y0 1, y0 3 d y dx 8 dx y0 4 19. 4y 4y y 0, y0 1, y0 1.5 S Click here for solutions. 16y 0 0. y 5y 3y 0, y0 1, 1. y 16y 0, y4 3,. y y 5y 0, y 0, 3. y y y 0, y0, 4. y 1y 36y 0, y1 0, 5 3 Solve the boundary-value problem, if possible. 5. 4y y 0, y0 3, 6. y y 0, y0 1, 7. y 3y y 0, y0 1, 8. y 100y 0, y0, 9. y 6y 5y 0, y0 1, 30. y 6y 9y 0, y0 1, 31. y 4y 13y 0, y0, 3. 9y 18y 10y 0, y0 0, 33. Let L be a nonzero real number. (a) Show that the boundary-value problem y y 0, y0 0, yl 0 has only the trivial solution y 0 for the cases and. (b) For the case, find the values of for which this problem has a nontrivial solution and give the corresponding solution. 0 0 0 y 4 y1 y4 4 y y0 1 y3 0 y 5 y0 4 y1 1 y y1 0 y 1 y 1 34. If a, b, and c are all positive constants and yx is a solution of the differential equation ay by cy 0, show that lim x l yx 0.

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 7 ANSWERS S Click here for solutions. 1. y c 1e 4x c e x 3. y e4x c 1 cos 5x c sin 5x 5. y c 1e x c xe x 7. y c 1 cosx c sinx 9. y c 1 c e x4 11. y c 1e (1s )t (1s )t c e 13. y e t [c 1 cos(s3t) c sin(s3t)] 15. 40 g f _0. 1 _40 All solutions approach 0 as x l and approach as x l. 17. y e 3x e x 19. y e x/ xe x 1. y 3 cos 4x sin 4x 3. y e x cos x 3 sin x 5. 7. y 3 cos( 1 x) 4 sin( 1 x) y e x3 e 3 1 9. No solution 31. y e x cos 3x e 33. (b), n a positive integer; y C sinnxl n L e x 1 e 3 sin 3x

8 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS SOLUTIONS 1. The auxiliary equation is r 6r +8=0 (r 4)(r ) = 0 r =4, r =. Then by (8) the general solution is y = c 1e 4x + c e x. 3. The auxiliary equation is r +8r +41=0 r = 4 ± 5i. Then by (11) the general solution is y = e 4x (c 1 cos 5x + c sin 5x). 5. The auxiliary equation is r r +1=(r 1) =0 r =1. Then by (10), the general solution is y = c 1e x + c xe x. 7. The auxiliary equation is 4r +1=0 r = ± 1 i,soy = c 1 cos 1 x + c sin 1 x. 9. The auxiliary equation is 4r + r = r(4r +1)=0 r =0, r = 1 4,soy = c1 + ce x/4. 11. The auxiliary equation is r r 1=0 r =1±,soy = c 1e (1+ )t + c e (1 )t. 13. The auxiliary equation is r + r +1=0 r = 1 ± 3 3 3 c i,soy = e t/ 1 cos t + c sin. t 15. r 8r +16=(r 4) =0so y = c 1e 4x + c xe 4x. The graphs are all asymptotic to the x-axis as x, and as x the solutions tend to ±. 17. r +5r +3=(r +3)(r +1)=0,sor = 3, r = 1 and the general solution is y = c 1e 3x/ + c e x.then y(0) = 3 c 1 + c =3and y 0 (0) = 4 3 c 1 c = 4,soc 1 =and c =1. Thus the solution to the initial-value problem is y =e 3x/ + e x. 19. 4r 4r +1=(r 1) =0 r = 1 and the general solution is y = c1ex/ + c xe x/.theny(0) = 1 c 1 =1and y 0 (0) = 1.5 1 c1 + c = 1.5,soc = and the solution to the initial-value problem is y = e x/ xe x/. 1. r +16=0 r = ±4i and the general solution is y = e 0x (c 1 cos 4x + c sin 4x) =c 1 cos 4x + c sin 4x. Then y π 4 = 3 c1 = 3 c 1 =3and y 0 π 4 =4 4c =4 c = 1,sothe solution to the initial-value problem is y =3cos4x sin 4x. 3. r +r +=0 r = 1 ± i and the general solution is y = e x (c 1 cos x + c sin x). Then=y(0) = c 1 and 1=y 0 (0) = c c 1 c =3and the solution to the initial-value problem is y = e x ( cos x +3sinx). 5. 4r +1=0 r = ± 1 i and the general solution is y = c1 cos 1 x + c sin 1 x.then3=y(0) = c 1 and 4 =y(π) =c, so the solution of the boundary-value problem is y =3cos 1 x 4sin 1 x. 7. r 3r +=(r )(r 1) = 0 r =1, r =and the general solution is y = c 1 e x + c e x.then 1=y(0) = c 1 + c and 0=y(3) = c 1 e 3 + c e 6 so c =1/(1 e 3 ) and c 1 = e 3 /(e 3 1). The solution of the boundary-value problem is y = ex+3 e 3 1 + ex 1 e. 3

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 9 9. r 6r +5=0 r =3± 4i and the general solution is y = e 3x (c 1 cos 4x + c sin 4x). But1=y(0) = c 1 and =y(π) =c 1e 3π c 1 =/e 3π, so there is no solution. 31. r +4r +13=0 r = ± 3i and the general solution is y = e x (c 1 cos 3x + c sin 3x). But =y(0) = c 1 and 1=y π = e π ( c ), so the solution to the boundary-value problem is y = e x ( cos 3x e π sin 3x). 33. (a) Case 1 (λ =0): y 00 + λy =0 y 00 =0which has an auxiliary equation r =0 r =0 y = c 1 + c x where y(0) = 0 and y(l) =0. Thus, 0=y(0) = c 1 and 0=y(L) =c L c 1 = c =0. Thus, y =0. Case (λ <0): y 00 + λy =0has auxiliary equation r = λ r = ± λ (distinct and real since λ<0) y = c 1 e λx + c e λx where y(0) = 0 and y(l) =0.Thus,0=y(0) = c 1 + c ( )and 0=y(L) =c 1e λl + c e λl ( ). Multiplying ( )bye λl andsubtracting( )givesc e λl e λl c 1 =0from ( ). Thus, y =0for the cases λ =0and λ<0. =0 c =0and thus (b) y 00 + λy =0has an auxiliary equation r + λ =0 r = ±i λ y = c 1 cos λx+ c sin λx where y(0) = 0 and y(l) =0.Thus,0=y(0) = c 1 and 0=y(L) =c sin λl since c 1 =0.Sincewe cannot have a trivial solution, c 6= 0and thus sin λl =0 λl = nπ where n is an integer λ = n π /L and y = c sin(nπx/l) where n is an integer.