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54 CHAPTER 2 Linear Second-Order Equations

2. If λ 1 = λ 2 ,
y(x) = c 1 e λ 1 x + c 2 xe λ 1 x .

3. If the roots are complex α ± iβ,
αx
αx
y(x) = c 1 e cos(βx) + c 2 e sin(βx).

SECTION 2.2 PROBLEMS

In each of Problems 1 through 10, write the general (c) Show that, as → 0, the solution in part (b) does
solution. not approach the solution in part (a), even though the
differential equation in part (b) would appear to more
1. y − y − 6y = 0

closely resemble that of part (a) as is chosen smaller.
2. y − 2y + 10y = 0

22. (a) Find the solution ψ of the initial value problem
3. y + 6y + 9y = 0

2

4. y − 3y = 0 y − 2αy + α y = 0; y(0) = c, y (0) = d

5. y + 10y + 26y = 0 with α = 0.

6. y + 6y − 40y = 0 (b) Find the solution ψ of

7. y + 3y + 18y = 0

2
2

y − 2αy + (α − )y = 0; y(0) = c, y (0) = d.

8. y + 16y + 64y = 0

9. y − 14y + 49y = 0 (c) Is it true that ψ (x) → ψ(x) as → 0?

10. y − 6y + 7y = 0 23. Suppose ϕ is a solution of
In each of Problems 11 through 20, solve the initial value y + ay + by = 0; y(x 0 ) = A, y (x 0 ) = B

problem.
with a,b, A,and B as given numbers and a and b

11. y + 3y = 0; y(0) = 3, y (0) = 6

positive. Show that

12. y + 2y − 3y = 0; y(0) = 6, y (0) =−2

13. y − 2y + y = 0; y(1) = y (1) = 0 lim ϕ(x) = 0.
x→∞

14. y − 4y + 4y = 0; y(0) = 3, y (0) = 5
24. Use power series expansions to derive Euler’s for-

15. y + y − 12y = 0; y(2) = 2, y (2) =−1

mula. Hint: Write

16. y − 2y − 5y = 0; y(0) = 0, y (0) = 3

∞ 1
17. y − 2y + y = 0; y(1) = 12, y (1) =−5 e = x ,

n
x
n!
18. y − 5y + 12y = 0; y(2) = 0, y (2) =−4 n=0

∞ n
19. y − y + 4y = 0; y(−2) = 1, y (−2) = 3 (−1) 2n+1

sin(x) = x ,
20. y + y − y = 0; y(−4) = 7, y (−4) = 1 n=0 (2n + 1)!

21. This problem illustrates how small changes in the and
coefficients of a differential equation may cause dra-
matic changes in the solution. (−1) n
∞
2n
(a) Find the general solution ϕ(x) of cos(x) = x .
(2n)!
n=0
2

y − 2αy + α y = 0
with α = 0. Let x = iβ with β real, and use the fact that
(b) Find the general solution ϕ (x) of
n
n
2n
i = (−1) and i 2n+1 = (−1) i.
2
2

y − 2αy + (α − )y = 0
with a positive constant. for every positive integer n.
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