Description
1. This problem pertains to the differential equation y�� + ω2y = sin ω0t, where ω �= 0 and ω0 is
close to but different from ω. sin ω0t (a) Verify that y1(t) = is a particular solution. ω2 − ω2
0
(b) As ω0 → ω show that one of the initial conditions y1(0) or y�(0) becomes infinite. 1
(c) Check that y2(t) = sin ω0t − sin ωt is the particular solution for which the initial conditions
ω2 − ω0
2
remain finite as ω0 → ω.
(d) By l’Hospital’s rule show that the limit as ω0 ω of y2(t) gives a particular solution of y�� +
ω2y = sin ωt.
→
2. Let f(x) and g(x) be two solutions of the differential equation y� = F(x, y) in a domain where
F satisfies the condition∗
:
y1 < y2 implies F(x, y2) − F(x, y1) ≤ L(y2 − y1).
Show that
|f(x) − g(x)| ≤ eL(x−a)
|f(a) − g(a)| if x > a.
3. Very that (sin x)/x, x satisfy the following equations, respectively, and thus obtain the second
solution.
(a) xy�� + 2y� + xy = 0 (x > 0),
(b) (2x − 1)y�� − 4xy� + 4y = 0 (2x > 1).
4. (a) Birkhoff-Rota, pp. 57, #4. (Typo. I(x) = q − p2/4 − p�
/2.)
(b) Birkhoff-Rota, pp. 57, #7(a). (Use part (a) instead of #6 as is suggested in the text.)
(c) Birkhoff-Rota, pp. 57, #7(b).
5. Let (cosh x)y�� + (cos x)y� = (1 + x2)y for a < x < b and let y(a) = y(b) = 1. Show that
0 < y(x) < 1 for a < x < b.
6. (a) Birkhoff-Rota, pp. 75, #3, (b) Birkhoff-Rota, pp. 75, #4.
∗It is called a one-sided Lipschitz condition.
1
.
