Nonlinear differential equation

Click For Summary

Homework Help Overview

The discussion revolves around a nonlinear differential equation given by d²x/dt² = -h*g/(h+x), which the original poster is struggling to solve due to their current level of understanding in high school mathematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various methods to approach the problem, including separation of variables and the use of the chain rule. There are suggestions to multiply both sides by dx/dt or to express the second derivative in terms of velocity (v = dx/dt) to facilitate integration.

Discussion Status

Some participants have offered guidance on potential methods for integration and transformation of the equation. There is an acknowledgment of the complexity involved due to the dependence of the right side on the unknown function x, indicating that multiple interpretations of the problem are being explored.

Contextual Notes

The original poster indicates a lack of familiarity with certain mathematical techniques, which may limit their ability to follow the discussion fully. There is also a mention of the need for explanations in the context of high school-level understanding.

mbadin
Messages
1
Reaction score
0

Homework Statement



In one problem I had got to this equations, but I was not able to solve it, because I'm actually
on high school.

The equation :
d^2/dt^2(x) = -h*g/(h+x)

I tried use separation of variables but I was not able to use the chain rule.
Can anybody show me the steps with explanation?
 
Last edited by a moderator:
Physics news on Phys.org
welcome to pf!

hi mbadin! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
mbadin said:
d2x/dt2 = -h*g/(h+x)

either multiply both sides by dx/dt

or use d2x/dt2 = vdv/dx

(where v = dx/dt … you can prove it using the chain rule :wink:)
 
OK, here is how your equation appears to me:
[tex]\frac{d^2x}{dt^2} = -\frac{hg}{h+x}[/tex]
If i understood your question correctly, then you need to integrate twice to get rid of the differentiation.
 
sharks said:
OK, here is how your equation appears to me:
[tex]\frac{d^2x}{dt^2} = -\frac{hg}{h+x}[/tex]
If i understood your question correctly, then you need to integrate twice to get rid of the differentiation.
It's not that easy because the right side, that you want to integrate with respect to t depends upon the unknown function x.

I would do what tiny-tim suggested: Let v= dx/dt so that [itex]d^2x/dt^2= dv/dt[/itex] and then, by the chain rule,
[tex]\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}[/tex]

Your equation becomes
[tex]v\frac{dv}{dx}= -\frac{hg}{h+ x}[/tex]
which can be integrated as
[tex]\int v dv= hg\int \frac{dx}{h+ x}[/tex]

It might well give you a function v= dx/dt that is difficult to integrate but that is the most direct method to solve this equation.
 

Similar threads

Replies
6
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K