Solve second order differential equation

Click For Summary

Discussion Overview

The discussion revolves around solving the second order differential equation \(\ddot{\phi} = -w^{2}\phi\) by direct substitution of the function \(\phi(t) = Asin(wt) + Bcos(wt)\). Participants explore the process of substitution and differentiation to verify if the proposed function is indeed a solution to the equation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about what is meant by "direct substitution" in the context of solving the differential equation.
  • Another participant questions whether "direct substitution" is different from normal substitution.
  • A participant suggests that the task is to show that the second derivative of the function equals the negative square of \(w\) times the function itself.
  • There is a clarification that the process involves substituting the function into the differential equation and performing differentiation twice to demonstrate that it reduces to an identity.
  • One participant expresses uncertainty about whether they are substituting or differentiating.
  • Another participant confirms that the goal is to differentiate the function and check if it equals \(-w^{2}\phi\).
  • There is a mention of the complexity of the derivative and a question about the treatment of \(wt\) as constants versus variables.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the clarity of the term "direct substitution" and the steps involved in the process. There is ongoing uncertainty regarding the differentiation and substitution process.

Contextual Notes

Some participants express confusion about the definitions and steps involved in substitution and differentiation, indicating a potential lack of clarity in the problem statement or understanding of the terms used.

Oblio
Messages
398
Reaction score
0
Solve by direct substitution taht the function [tex]\phi[/tex](t) = Asin(wt) + Bcos(wt) is a solution of the second order differential equation [tex]\ddot{\phi}[/tex] = -w[tex]^{2}[/tex][tex]\phi[/tex]. ( Since this solution involves two arbitrary constants - the coeffecients of the sine and consine functions - it is in fact the general solution).

I'm not sure what is meant by direct substitution..
Can someone help me get started?
Thanks a lot!
 
Physics news on Phys.org
What does the word "substitution" normally mean?
 
Is direct substitution different then normal?
I tried subbing in Asin(wt) + Bcos(wt) into the second order differential equation but that got me nowhere...
 
Well, you are asked to show that the second derivative of the given function equals the negative square of w times the given function.

Have you tried to show that?
 
That's why I'm confsued.
Am i substituting or differentiating?
 
Eeh?

You are to show that by SUBSTITUTING into the diff.eq, which requires you to perform differentiation twice, the differential equation reduces to to an IDENTITY, valid for all t, i.e, that the given function is, indeed, a SOLUTION to the differential equation.
 
So I am differentiating

[tex]\ddot{\phi}[/tex]= -w[tex]^{2}[/tex](Asin(wt) + Bcos(wt))
 
Quite so!

Does that equal [itex]-w^{2}\phi[/itex]??
 
I hope so, it probably will turn out not to be so when I do it wrong :)
 
  • #10
(long derivative..)
 
  • #11
wt wouldn't be constants... not treated as normal variables?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 52 ·
2
Replies
52
Views
9K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K