• Support PF! Buy your school textbooks, materials and every day products Here!

Struggling with a second-order differential equation

  • Thread starter tomfitzyuk
  • Start date
  • #1
15
0
Hey,

For my Physics coursework I'm trying to show how the equation to describe simple harmonic motion, this is where I'm up to now:
http://img408.imageshack.us/img408/4153/untitled8ph.jpg [Broken]

x displacement;
t is time;
k is spring constant;
m is mass.

A is an unknown constant;
b is an unknown constant.

Could somebody check through my working, see if I've gone wrong and hopefully point me in the right direction as to what to do (I've tried respresting the function of e in terms of trigonometric functions but this never made sense.)

Thanks,
Tom

PS. I know it's possible, from the differential equation given, to solve it straight off using the method by assuming it's a quadratic (or something similar) but I want this proof to be rigourous and therefore everything (to some degree) must be explained by me.
 
Last edited by a moderator:

Answers and Replies

  • #2
Integral
Staff Emeritus
Science Advisor
Gold Member
7,200
56
Have you checked the solution? Does it satisfy the original DE?

EDIT: Just realized that you do not have a complete statement of the DE. What are your initial conditions?
 
Last edited:
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
955
You have correctly determined that [itex]Ae^{\sqrt{\frac{k}{m}}it}[/itex]
and [itex]Be^{-/sqrt{\frac{k}{m}}it}[/itex] are solutions to the differential equation. However, you have no reason to assert that
[tex]Ae^{\sqrt\frac{k}{m}}it}-A\frac{k}{m}e^{-\sqrt\frac{k}{m}}it}= 0[/tex]. Use those two solutions to form the general solution to the differential equation. Do you know how to do that?

Also do you know that [itex]e^{ix}= cos(x)+ i sin(x)[/itex]?
 
  • #4
15
0
Just realized that you do not have a complete statement of the DE. What are your initial conditions?
Not sure what you mean by initial conditions and having a complete statement of the DE.

HallsofIvy said:
You have correctly determined that [itex]Ae^{\sqrt{\frac{k}{m}}it}[/itex]
and [itex]Be^{-/sqrt{\frac{k}{m}}it}[/itex] are solutions to the differential equation. However, you have no reason to assert that
[tex]Ae^{\sqrt\frac{k}{m}}it}-A\frac{k}{m}e^{-\sqrt\frac{k}{m}}it}= 0[/tex]. Use those two solutions to form the general solution to the differential equation. Do you know how to do that?
Also do you know that [itex]e^{ix}= cos(x)+ i sin(x)[/itex]?
I don't know what you mean by the general solution the differential equation, I haven't started University and my knowledge of differential equations is limited to first-order differential equations.

I know that [itex]e^{ix} = cos(x) + i sin(x)[/itex] which I've tried to use but failed. I end up with an imaginary part which doesn't make sense.

Thanks.
Tom.
 
  • #5
Initial conditions are usually given values of a function evaluated at a specific point.
For example, given:
y(0) = 3pi
y'(0) = 2
That would transform your problem into an initial-value problem, and you'll end up with a solution particular, i.e. satisfying specifically, to/for your differential equation.

A general solution is pretty much a solution for your differential equation, but not limited to that differential equation; i.e. it could have constants, like the A and B, and thus satisfy many differential equations.

I suggest you read up on http://tutorial.math.lamar.edu/AllBrowsers/3401/3401.asp [Broken] if you feel that your knowledge is somewhat lacking.
 
Last edited by a moderator:

Related Threads on Struggling with a second-order differential equation

Replies
4
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
1
Views
739
  • Last Post
Replies
3
Views
881
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
2
Views
625
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
1K
Replies
5
Views
859
  • Last Post
Replies
1
Views
366
Top