Differential Equation Help: As t approaches 0, y approaches

Click For Summary
The discussion revolves around solving a differential equation and finding the critical value a0 as t approaches 0. The correct solution for part (b) is identified as y = -cos(t)/(t^2) + (a*pi^2)/(4t^2) with a0 being 4/pi^2. There is uncertainty regarding the behavior of y as t approaches 0, with an initial assumption that y approaches 0 being deemed incorrect. Participants discuss the need for clarification on the Taylor expansion of cos(t) to aid in further understanding the problem. The exchange emphasizes the importance of precise mathematical expansions in solving differential equations.
TheCarl
Messages
21
Reaction score
0

Homework Statement



http://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest1886032entrance1_N10020.mml?size=14&rnd=1360201586591

(b) Solve the initial value problem and find the critical value a0 exactly.
y = ?​
a0 = ?​
(c) Describe the behavior of the solution corresponding to the initial value a0.
y -> ? as t -> 0​

The Attempt at a Solution



I got part b correct but I thought I'd put it in here to help speed the process for whoever can help me.

(b) y= -cos(t)/(t^2) + (a*pi^2)/(4t^2)

a0 = 4/pi^2

(c) I would think y would approach 0 as t approaches 0 but that apparently is wrong. This is where I need assistance. Any help is greatly appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
TheCarl said:
(b) y= -cos(t)/(t^2) + (a*pi^2)/(4t^2)
a0 = 4/pi^2
I assume a = a0, so y= (1-cos(t))t-2
Do you know an expansion for cos(t) valid in the vicinity of 0?
 
I apologize if I seem a bit dense but could you elaborate on your question about the expansion on cos(t)? I'm not entirely sure what you're asking.
 
Taylor expansion? Power series?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 2 ·
Replies
2
Views
920
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
1K
Replies
3
Views
2K
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K