SECTION 5.1

INTRODUCTION: SECOND-ORDER LINEAR EQUATIONS

In this section the central ideas of the theory of linear differential equations are introduced andillustrated concretely in the context of second-order equations. These key concepts includesuperposition of solutions (Theorem 1), existence and uniqueness of solutions (Theorem 2),linear independence, the Wronskian (Theorem 3), and general solutions (Theorem 4). Thisdiscussion of second-order equations serves as preparation for the treatment of nth order linearequations in Section 5.2. Although the concepts in this section may seem somewhat abstract tostudents, the problems set is quite tangible and largely computational.

In each of Problems 1-16 the verification that y1 and y2 satisfy the given differential equationis a routine matter. As in Example 2, we then impose the given initial conditions on the generalsolution y = c1y1 + c2y2. This yields two linear equations that determine the values of theconstants c1 and c2.

1. Imposition of the initial conditions (0) 0, (0) 5y y′= = on the general solution

1 2( ) x xy x c e c e−= + yields the two equations 1 2 1 20, 0c c c c+ = − = with solution

1 25 / 2, 5 / 2.c c= = − Hence the desired particular solution is y(x) = 5(ex - e-x)/2.

2. Imposition of the initial conditions (0) 1, (0) 15y y′= − = on the general solution3 3

1 2( ) x xy x c e c e−= + yields the two equations 1 2 1 21, 3 3 15c c c c+ = − − = with solution

1 22, 3.c c= = Hence the desired particular solution is y(x) = 2e3x - 3e-3x.

3. Imposition of the initial conditions (0) 3, (0) 8y y′= = on the general solution

1 2( ) cos 2 sin 2y x c x c x= + yields the two equations 1 23, 2 8c c= = with solution

1 23, 4.c c= = Hence the desired particular solution is y(x) = 3 cos 2x + 4 sin 2x.

4. Imposition of the initial conditions (0) 10, (0) 10y y′= = − on the general solution

1 2( ) cos5 sin 5y x c x c x= + yields the two equations 1 210, 5 10c c= = − with solution

1 23, 4.c c= = Hence the desired particular solution is y(x) = 10 cos 5x - 2 sin 5x.

5. Imposition of the initial conditions (0) 1, (0) 0y y′= = on the general solution2

1 2( ) x xy x c e c e= + yields the two equations 1 2 1 21, 2 0c c c c+ = + = with solution

1 22, 1.c c= = − Hence the desired particular solution is y(x) = 2ex - e2x.

6. Imposition of the initial conditions (0) 7, (0) 1y y′= = − on the general solution2 3

1 2( ) x xy x c e c e−= + yields the two equations 1 2 1 27, 2 3 1c c c c+ = − = − with solution

1 24, 3.c c= = Hence the desired particular solution is y(x) = 4e2x + 3e-3x.

7. Imposition of the initial conditions (0) 2, (0) 8y y′= − = on the general solution

1 2( ) xy x c c e−= + yields the two equations 1 2 22, 8c c c+ = − − = with solution

1 26, 8.c c= = − Hence the desired particular solution is y(x) = 6 − 8e-x.

8. Imposition of the initial conditions (0) 4, (0) 2y y′= = − on the general solution3

1 2( ) xy x c c e= + yields the two equations 1 2 24, 3 2c c c+ = = − with solution

1 214 / 3, 2 / 3.c c= = Hence the desired particular solution is y(x) = (14 - 2e3x)/3.

9. Imposition of the initial conditions (0) 2, (0) 1y y′= = − on the general solution

1 2( ) x xy x c e c x e− −= + yields the two equations 1 1 22, 1c c c= − + = − with solution

1 22, 1.c c= = Hence the desired particular solution is y(x) = 2e-x + xe-x.

10. Imposition of the initial conditions (0) 3, (0) 13y y′= = on the general solution5 5

1 2( ) x xy x c e c x e= + yields the two equations 1 1 23, 5 13c c c= + = with solution

1 23, 2.c c= = − Hence the desired particular solution is y(x) = 3e5x - 2xe5x.

11. Imposition of the initial conditions (0) 0, (0) 5y y′= = on the general solution

1 2( ) cos sinx xy x c e x c e x= + yields the two equations 1 1 20, 5c c c= + = with solution

1 20, 5.c c= = Hence the desired particular solution is y(x) = 5exsin x.

12. Imposition of the initial conditions (0) 2, (0) 0y y′= = on the general solution3 3

1 2( ) cos 2 sin 2x xy x c e x c e x− −= + yields the two equations 1 1 22, 3 2 5c c c= − + = with

solution 1 22, 3.c c= = Hence the desired particular solution is y(x) =e-3x(2 cos 2x + 3 sin 2x).

13. Imposition of the initial conditions (1) 3, (1) 1y y′= = on the general solution2

1 2( )y x c x c x= + yields the two equations 1 2 1 23, 2 1c c c c+ = + = with solution

1 25, 2.c c= = − Hence the desired particular solution is y(x) = 5x - 2x2.

14. Imposition of the initial conditions (2) 10, (2) 15y y′= = on the general solution2 3

1 2( )y x c x c x−= + yields the two equations 1 2 1 24 / 8 10, 4 3 /16 15c c c c+ = − = with

solution 1 23, 16.c c= = − Hence the desired particular solution is y(x) = 3x2 - 16/x3.

15. Imposition of the initial conditions (1) 7, (1) 2y y′= = on the general solution

1 2( ) lny x c x c x x= + yields the two equations 1 1 27, 2c c c= + = with solution

1 27, 5.c c= = − Hence the desired particular solution is y(x) = 7x - 5x ln x.

16. Imposition of the initial conditions (1) 2, (1) 3y y′= = on the general solution

1 2( ) cos(ln ) sin(ln )y x c x c x= + yields the two equations 1 22, 3.c c= = Hence thedesired particular solution is y(x) = 2 cos(ln x) + 3 sin(ln x).

17. If /y c x= then 2 2 2 2 2/ / ( 1) / 0y y c x c x c c x′ + = − + = − ≠ unless either c = 0or c = 1.

18. If 3y cx= then 3 2 4 46 6 6yy cx cx c x x′′ = ⋅ = ≠ unless 2 1.c =

19. If 1y x= + then 2 3/ 2 1/ 2 2 3/ 2( ) (1 )( / 4) ( / 2) / 4 0.yy y x x x x− − −′′ ′+ = + − + = − ≠

20. Linearly dependent, because

f(x) = π = π(cos2x + sin2x) = πg(x)

21. Linearly independent, because x x x3 2= + if x > 0, whereas x x x3 2= − if x < 0.

22. Linearly independent, because 1 1+ = +x c x( ) would require that c = 1 with x = 0,but c = 0 with x = -1. Thus there is no such constant c.

23. Linearly independent, because f(x) = +g(x) if x > 0, whereas f(x) = -g(x) if x < 0.

24. Linearly dependent, because g(x) = 2 f(x).

25. f(x) = exsin x and g(x) = excos x are linearly independent, because f(x) = k g(x)would imply that sin x = k cos x, whereas sin x and cos x are linearly independent.

26. To see that f(x) and g(x) are linearly independent, assume that f(x) = c g(x), and thensubstitute both x = 0 and x = π/2.

27. Let L[y] = y″ + py′ + qy. Then L[yc] = 0 and L[yp] = f, so

L[yc + yp] = L[yc] + [yp] = 0 + f = f.

28. If y(x) = 1 + c1cos x + c2sin x then y′(x) = -c1sin x + c2cos x, so the initial conditionsy(0) = y′(0) = -1 yield c1 = -2, c2 = -1. Hence y = 1 - 2 cos x - sin x.

29. There is no contradiction because if the given differential equation is divided by x2 toget the form in Equation (8) in the text, then the resulting functions p(x) = -4/x andq(x) = 6/x2 are not continuous at x = 0.

30. (a) y x23= and y x2

3= are linearly independent because x c x3 3= would requirethat c = 1 with x = 1, but c = -1 with x = -1.

(b) The fact that W(y1, y2) = 0 everywhere does not contradict Theorem 3, becausewhen the given equation is written in the required form

y″ - (3/x)y′ + (3/x2)y = 0,

the coefficient functions p(x) = -3/x and q(x) = 3/x2 are not continuous at x = 0.

31. W(y1, y2) = -2x vanishes at x = 0, whereas if y1 and y2 were (linearly independent)solutions of an equation y″ + py′ + qy = 0 with p and q both continuous on an openinterval I containing x = 0, then Theorem 3 would imply that W ≠ 0 on I.

32. (a) W = y1y2′ - y1′y2, so

AW' = A(y1′y2′ + y1y2″ - y1″y2 - y1′y2′)

= y1(Ay2″) - y2(Ay1″)

= y1(-By2′ - Cy2) - y2(-By1′ - Cy1)

= -B(y1y2′ - y1′y2)

and thus AW' = -BW.

(b) Just separate the variables.

(c) Because the exponential factor is never zero.

In Problems 33-42 we give the characteristic equation, its roots, and the corresponding generalsolution.

33. 2 3 2 0; 1, 2;r r r− + = = y(x) = c1ex + c2e

2x

34. 2 2 15 0; 3, 5;r r r+ − = = − y(x) = c1e-5x +c2e

3x

35. 2 5 0; 0, 5;r r r+ = = − y(x) = c1 + c2e-5x

36. 22 3 0; 0, 3 / 2;r r r+ = = − y(x) = c1 + c2e-3x/2

37. 22 2 0; 1, 1/ 2;r r r− − = = − y(x) = c1e-x/2 + c2e

x

38. 24 8 3 0; 1/ 2, 3 / 2;r r r+ + = = − − y(x) = c1e-x/2 +c2e-

3x/2

39. 24 4 1 0; 1/ 2, 1/ 2;r r r+ + = = − − y(x) = (c1 + c2x)e-x/2

40. 29 12 4 0; 2 / 3, 2 / 3;r r r− + = = − − y(x) = (c1 + c2x)e2x/3

41. 26 7 20 0; 4 / 3, 5 / 2;r r r− − = = − y(x) = c1e-4x/3 + c2e

5x/2

42. 235 12 0; 4 / 7, 3 / 5;r r r− − = = − y(x) = c1e-4x/7 + c2e

3x/5

In Problems 43-48 we first write and simplify the equation with the indicated characteristic roots,and then write the corresponding differential equation.

43. 2( 0)( 10) 10 0;r r r r− + = + = 10 0y y′′ ′+ =

44. 2( 10)( 10) 100 0;r r r− + = − = 100 0yy′′ − =

45. 2( 10)( 10) 20 100 0;r r r r+ + = + + = 20 100 0y y y′′ ′+ + =

46. 2( 10)( 100) 110 1000 0;r r r r− − = − + = 110 1000 0y y y′′ ′− + =

47. 2( 0)( 0) 0;r r r− − = = 0y′′ =

48. 2( 1 2)( 1 2) 2 1 0;r r r r− − − + = − − = 2 0y y y′′ ′− − =

49. The solution curve with y y( ) , ( )0 1 0 6= ′ = is y x e ex x( ) = −− −8 7 2. We find that

′ =y x( ) 0 when x = ln(7/4) so e ex x− −= =4 7 16 492/ /and . It follows thaty(ln( / )) /7 4 16 7= , so the high point on the curve is (ln( / )), / ) ( . , . )7 4 16 7 0 56 2 29≈ ,which looks consistent with Fig. 5.1.6.

50. The two solution curves with y a y b( ) ( )0 0= =and (as well as ′ =y ( )0 1) are

y a e a e

y b e b e

x x

x x

= + − +

= + − +

− −

− −

( ) ( ) ,

( ) ( ) .

2 1 1

2 1 1

2

2

Subtraction and then division by a - b gives 22e ex x− −= , so it follows that x = -ln 2.

Now substitution in either formula gives y = -2, so the common point of intersection is(-ln 2, -2).