7.

2 The Auxiliary Equation: Distinct Roots 119

The auxiliary equation is

m 2 -4 = 0,
with roots m = 2, - 2. Hence the general solution of the differential equation is
x = c] e
21
+ C2e - 2t.
It remains to enforce the conditions at t = O. Now
dx
- = 2c]e 21 - 2c?e- 21.
dt -
Thus the condition that x = 0 when t = 0 requires that
0= CI + C2,
and the condition that dx / dt = 3 when t = 0 requires that
3 = 2c] - 2C2.

From the simultaneous equations for CI and C2 we conclude that CI 1

4
and
C2 = - t.
Therefore,
x = t(e 21 - e- 21) ,
which can also be put in the form
x = ~ sinh (2t).

• Exercises
•
In Exercises 1 through 22, find the general solution. When the operator D is used, itis implied
that the independent variabl e is x .

l. (D 2 + 2D - 3)y = O. 10. (4D 313D - 6)y = O.

-

2. (D 2 + 2D)y = O. d x d 2x
3
dx
11. -+ - 2- 2-=0.
3. (D 2 + D - 6) y = O. 3
dt dt dt
4. (D 2 - 5D + 6)y = O. d 3x dx
12. - 3 - 19- + 30x = O.
5. (D 3 + 3D 2 - 4D)y = O. dt dt
6. CD 3 - 3D 2 - lOD)y = O. 13. (9D 3 -7D+2)y=O.
7. (D 3 + 6D 2 + lID + 6)y = O. 14. (4D 3 - 2ID - lO)y = O.
8. (D 3 + 3D 2 - 4D - 12)y = O. 15. CD 3 - 14D + 8)y = O.
9. (4D 3 - 7 D + 3)y = O. 16. (D 3 - D2 - 4D - 2)y = O.
17. C4D4 - 8D 3 - 7D 2 + llD + 6)y = O.
18. (4D4 - I6D 3 + 7D2 + 4D - 2)y = O.
19. (4D 4 + 4D 3 - 13D 2 -7D + 6)y = O.
120 Chapter 7 Linear Equations with Constan.t Coefficients

20 . (4D 5 - 8D 4 - 17 D3 + 12D2 + 9D)y = 0.

21. (D 2 - 4aD + 3( 2 )y = 0; a real "# 0.
22. [D 2 - (a + b)D + ab]y = 0; a and b real and unequal.

In Exercises 23 and 24, find the particular solution indicated.

23. (D 2 - 2D - 3)y = 0; when x = 0, y = 0, yt = - 4.

24. (D 2 - D - 6) y = 0; when x = 0, y = 0, and when x = 1, y = e 3 .

In Exercises 25 through 29, find for x = I the y value for the particular solution required.

25. (D 2 - 2D - 3) y = 0; when x = 0, y = 4, y' = 0.

26. CD 3 - 4D)y = 0; when x = 0, y = 0 , yt = 0, y" = 2.
27. (D 2 - D - 6)y = 0; when x = 0, y = 3, yt = - 1.
28. CD2 + 3D - 10)y = 0; when x = 0, y = 0, and when x = 2 , y = l.
29. CD 3
- 2D2 - 5D + 6) y = 0; when x = 0, y = 1, yt = - 7, y" = - 1.

7 .3 The Auxiliary Equation : Repeated Roots

Suppose that in the eq uation

f(D) y = ° (1)

the operator feD) has repeated factors; that is, the auxiliary equation f(m) =
has repeated roots. Then the method of the preceding section does not yield
°
the general solution. Let the auxiliary equation have three equal roots m I = b,
m 2 = b, m 3 = b. The correspo nding part of the solution yielded by the method
of Section 7.2 is
y = C Ie + C2evx + C3evx,
Vx

vx (2)
y = (CI + C2 + c3)e .

Now (2) can be replaced by

(3)

with C4 = CI + C2 + C3. Thus, corre ponding to the three roots under consideration ,
this method has yielded only the solution (3). The difficulty is present, of course,
because the three solutions corresponding to the roots m I = m2 = m 3 = bare
not linearly independent.
What is needed is a method for obtaining n linearly independent solutions cor-
responding to n equal roots of the auxili ary equation. Suppose that the auxiliary
°
eq uation f(m) = has the n roots
122 Chapter 7 Linear Equations with Constanl Coefficients

EXAMPLE 7.4
Solve the equation

(D 4 - 7D3 + 18D 2 - 20D + 8)y = O. (10)

With the aid of synthetic division, it is easily seen that the auxiliary equation

m 4 - 7m 3 + 18m 2 - 20m +8= 0

has the roots m = 1, 2 , 2, 2. Then the general sol ution of equation (10) is

or

y= Cte x + ( C2 + C3X + C4X 2) e 2r

' .

•
EXAMPLE 7.5
Solve the equation

d4 y d 3 y d2 y
-+2-+-= 0.
dx 4 dx 3 dx 2
The auxiliary equation is

m
4
+ 2m 3 + m 2 = 0,
with roots m = 0, 0, - 1, - 1. Hence the desired solution is

• Exercises
•
In Exercises 1 through 20, fi nd the general solutj on.

1. ( D 2 - 6D +9)y = 0 . 4. (D 3 8D 2 + 16D)y = O.
-

2. (D 2 + 4D + 4) y = O. 5. (D4 + 6D 3 + 9D2)y = O.
3. (4D 3 + 4D2 + D)y = O. 6. (D 3 - 3D 2 + 4)y = O.
3
7. (4D - 3D + Oy = O.
8. ( D4 - 3D 3 - 6D 2 + 28D - 24)y = O.
9. (D 3 + 3D 2 + 3D + l) y = O. 11. (D s - D 3 )y = O.
10. (D 3 + 6D 2 + 12D + 8)y = O. 12. (D s - 16D 3 )y = O.
13. 4 2
(4D +4D3 - 3D - 2D + l)y = O.
 7.4 A Definition of exp z forlmaginary z 123

14. (4D 4 - 4D 3 - 23D 2 + 12D + 36)y = 0.

15. (D 4 + 3D 3 - 6D 2 - 28D - 24)y = 0.
16. (27D4 - 18D 2 + 8D - ' l)y = 0.
17. (4D 5 - 23D 3 - 33D 2 - 17D - 3)y = 0.
18. (4D 5 - 15D 3 - 5D2 + 15D + 9)y = 0.
19. (D 4 - 5D2 - 6D - 2)y = 0.
20. (D 5 - 5D 4 + 7 D3 + D2 - 8D + 4)y = 0.

In Exercises 2 J through 26, find the particular olution indicated.

2l. (D 2 +4D+4)y = 0; when x = 0, y = 1, y' =-1.

22. The equation of Exercise 21 with the condition that the graph of the solution
pass through the points (0, 2) and (2, 0).
23. (D 3 - 3D - 2)y = 0; when x = 0, y = 0, y' = 9, y" = 0.
24. (D 4 + 3D 3 + 2D2)y = 0; when x = 0, y = 0, y' = 4, y" = - 6, y"l = 14.
25. The equation of Exercise 24 with the conditions that when x = 0, y = 0,
y' = 3, y" = - 5, y"' = 9.
26. CD 3 + D2 - D - l)y = 0; when x = 0, y = 1, when x = 2, y = 0, and
also with the condition as x -7 00, Y -7 0.

In Exercises 27 through 29, find for x = 2 the y value for the particular solution required.
27. (4D 2 - 4D + l)y = 0; when x = 0, y = - 2, y' = 2.
28. (D 3 + 2D2)y = 0; when x = 0, y = -3, y' = 0, y" = 12.
29. CD 3 + 5D2 + 3D - 9)y = 0; when x = 0, y = -1, when x = 1, y = 0,
and also with the condition as x -7 00, Y -7 0.

7.4 A Definition of exp z for Imaginary z

Since the auxiliary equation may have imaginary roots, we need to lay down a
definition of exp z for imaginary z.
Let z = 0: + if3 with 0: and f3 real. Since it is desirable to have the ordinary
laws of exponents remain valid, it is wise to require that

(l)

To eCL with 0: real , we attach the usual meaning.

Now consider e i /3, f3 real. In calculus it is shown that for all real x

X x2 x3 x"
eX = 1 + - + - + - + ... + - + .. . (2)
I! 21 31 n!
130 Chapter 7 Linear Equations with COl/stant Coefficients

The initial conditions now require that °= c, and 2 = 2C2, so finally,

y = sin h 2x.
Note that if we were to choose the alternative form

for the general solution of (7), we would obtain the same result with a little more
fuss in detel111ining C3 and C4 . Indeed, one major reaso n for using the hyperbolic
°
fu nctions is that cosh ax and sinh ax have values I and when x = 0, a fact that
is particularly useful in solving initial value problems .

• Exercises
•
Find the general sol ution except when the exercise stipulates otherwise.

1. Verify directly that the relation

(A)

satisfies the equation

[CD - a)2 + b2Jy = 0.

2. (D 2 - 2D + 5)y = O. 7. CD2 - 4D + 7)y = O.

3. (D 2 - 2D + 2)y = O. S. CD 3 + 2D2 + D + 2)y = O.
4. (D 2 + 9)y = O. 9. (D 4 + 2D 3 + 10D2)y = O.
S. (D 2 - 9)y = O. 10. (D 4 -2D 3 +2D 2 - 2D+ l )y = O.
6. (D 2 + 6D + 13)y = O. 11. (D4 + lSD 2 + Sl)y = O.
12. (2D 4 + J ID J - 4D2 - 69D + 34)y = O.
13. (D + 9D + 24D2 + 16)y = O.
6 4

14. (2D 3 - D2 + 36D - lS)y = O.

15. (D 2 -l)y = 0; when x = 0, y = Yo, y' = O.
16. (D 2 + J)y = 0; when x = 0, y = yo, y' = O.
17. (D J +7D 2 + J9D + 13)y = 0; when x = 0, y = 0, y' = 2, y" = -12.
IS. (D 5 + D4 - 7D 3 - llD2 - SD - 12)y = O.
d 2x dx
19. -2 +k 2 x = 0, k real; when t = 0, x = 0, - = Va.
dt dt
20. (D + D2 + 4D + 4)y = 0; when x = 0, y = 0, y' = -1, y" = S.
J

czZx dx 2 . dx
21. dt 2 + 2b di + k x = 0, k > b > 0, when t = 0, x = 0, dt = Vo.
 7.6 A NOie on H yperbolic FUIlC/ioli s 131

• Miscellaneous Exercises
Obtai n the general solution unless otherwise instructed.

1. (D 2 +3D)y=0. 10. (4D 3 -2 1D - IO )y=0.

2. (9D 4 + 6D 3 + D2)y = O. 11. (4D 3 - 7 D + 3)y = O.
3. (D 2 + D - 6)y = O. 12. (D J - 14D + 8)y = O.
4. (D J + 2D2 + D + 2)y = O. 13. (8D 3 - 4D2 - 2D + l )y = O.
3 2
S. (D - 3D + 4)y = O. 14. (D4 + D3 - 4D2 - 4D)y = O.
6. (D - 2D2 - 3D)y = O.
3
IS . (D 4-2D 3+SD 2-8D+4)y = O.
7. (4D 3 - 3D + l )y = O. 16. (D 4 + 2D2 + I)y = O.
8. (D J + 3D 2 - 4D - 12»)1 = O. 17. (D 4 + SD 2 + 4)y = O.
2
9. (D + 3D + 3D + 1))1 = O.
3
18. (D 4 + 3D J - 40)y = O.
19. (D 4 - 11D 3 + 36D 2 - 16D - 64)y = O.
20. (D 2 + 2D + S) y = O.
21. (D 4 + 4D 3 + 2D2 - 8D - 8)y = O.
22. (4D4 - 24D 3 + 3SD 2 + 60 - 9»)1 = O.
23. (4D 4 + 20D 3 + 350 2 + 2SD + 6)y = O.
24. (04 - 7D3+ IID 2 +SD - 14)y=0 .
25. (D 3 +S D 2 +7D+3)y=0. 33 . (D 4 -D 3 - 3D 2 +D+2)y=0.
26. (D - 2D2 + D - 2)y = O.
3 34. (D 3 - 2D2 - 3D + 10) y = o.
27. (D J - D2 + D - I)y = O. 35. (D s + D4 - 6D 3 )y = O.
28. (D J +40 2 +SD)y =0. 36. (4D 3 +28D 2 +6ID+37)y=0.
29. (D4 - 13D 2 + 36)y = O. 37 . (4DJ+ 12D2+ 13D + 10)y = o.
30. (D4-SDJ+SD2 +S0 -6)y = O. 38. (18D 3 - 33D 2 +20D -4)y = O.
31. (4D 3 + 8D 2 - 11 D + 3»)1 = O. 39. (D s -2D 3 - 2D 2-3D-2)y = O.
32. (D J + D2 - 160 - 16)y = O. 40. (D 4-2D 3 +2D 2-2D+ 1).'1 = O.
41. (D 5 - ISD 3 + lOD 2 + 60D - 72)y = O.
42. (4D4 - ISD 2 + SD + 6) y = O.
43. (D 4 + 3D 3 - 6D 2 - 28D - 24)y = O.
44 . (4D 4 - 4D 3 - 23 D 2 + 12D + 36)y = O.
45 . (4D 5 - 23D J - 33D 2 - 17 D - 3)y = O.
46. (D 2 - D - 6) y = 0; when x = 0, y = 2, )I' = l.
47. (D4 + 6D 3 + 9D 2 »)I = 0; when x = 0, y = 0 , y' = 0, y" = 6, and as
x ~ 00, y' ~ 1. For this particular solution, find the value of y when
x =1.
132 Chapter 7 Linear Equations with Constant Coefficienls

48. (D 3 + 6D 2 + 12D + 8)y = 0; when x = 0, y = 1, y' = - 2, y" = 2.

49. (D s + D4 - 9D 3 - 13D 2 + 8D + 12) y = O.
50. (4D 5 + 4D4 - 9D 3 - llD2 + D + 3)y = O.
51. (D s + D4 - 7 D3 - J ID2 - 8D - 12)y = O.

I 7.7 I Computer Supplement

The techniques described in the Computer Supplement to Chapter 2 extend easily
to higher-order equations. We can ill ustrate this with Example 7.6 in Section 7.5

(D 3 - 3D 2 +9D + 13»), = O.

If we add the initial conditions, yeO) = 1, y' (0) = 2, y" (0) = 3, Maple solves
the problem with the commands
>di f f(y (x) ,x$3 )-3* di f f(y (x) ,x$2)
+9*diff(y(x) ,x)+13*y(x)=O;
d3 d2 d
dx 3 y(x) - 3 dx 2 y(x) + 9 dx y(x) + 13 y(x) = 0

>dso lve({ " ,y(O) = l, D (y ) (O) =2,D(D(y)) (O ) =3 },y(x));

4 e- x S e2 Xcos(3x) 4e 2 x sin(3x)
y (x) = - 9- + 9 + 9

We can also use Maple to plot the resulting sol ution with the command
>p l ot (rhs(" ) , x=-2 .. 2);
See Figure 7.2.

20

15

10

-2 -1 0
-5

- 10

Figure 7.2