Second Order Differential Equation - Can't solve

  • Thread starter jumbogala
  • Start date
  • #1
423
2

Homework Statement


The equation is
x'' - m2x - m2b = 0.

m and b are constants. x'' is the second derivative of x, with respect to time.

Homework Equations





The Attempt at a Solution


When I did a differential equations course, I was taught to find the characteristic equation of the differential equation, then solve that. I tried it, but obviously it gave me an answer that was e to the power of something.

Which is not in the right format. How did the book do this? Can anyone get me started?
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
619
x=a+Acos(wt + B) doesn't solve that equation. It does solve x''+ w^2*x-w^2*a=0. Is there a typo?
 
  • #3
423
2
Oh sorry, that negative should be a positive. The equation you gave is the correct one.
 
  • #4
Dick
Science Advisor
Homework Helper
26,263
619
Oh sorry, that negative should be a positive. The equation you gave is the correct one.

If it's positive then the real solutions to x''+w^2*x=0 aren't exponentials, are they?
 
  • #5
423
2
They aren't? (Sorry it's been a long time since I did differential equations. I can't remember methods to solve very well.)

Ok so I am remembering now that I have to find a solution to the homogeneous equation x'' + w^2*x = 0 first. So this has complex roots?

My old notes show that the solution should be in the form of e multiplied by a cos or sin still though.
 
  • #6
Dick
Science Advisor
Homework Helper
26,263
619
D^2+w^2=0 has complex roots. D=iw and -iw. So complex solutions are e^(iwt) and e^(-iwt). What kind of real solution do those produce? Is it coming back yet?
 
  • #7
423
2
You can use euler's identity: e^(iwt) = cos(wt) + isin(wt)

So I could try a general solution of that form. Like Acos(wt) + Bsin(wt) and use that.

The book's solution was x=a+Acos(wt + B), so if A and a are constants of integration this should work. I'm not sure where B comes from though, or if I have to include the i in front of the sin in my general solution.
 
  • #8
2,981
5
It is an inhomogeneous equation. It can be rewritten as:

[tex]
x'' - m^{2} \, x = m^{2} b
[/tex]

The corresponding homogeneous equation is:

[tex]
x'' - m^{2} \, x = 0
[/tex]

What is the general solution of this equation?

A particular solution of the inhomogeneous equation is a constant:

[tex]
x_{p}(t) = A
[/tex]

What value of A satisfies the equation?
 
  • #9
Dick
Science Advisor
Homework Helper
26,263
619
You can use euler's identity: e^(iwt) = cos(wt) + isin(wt)

So I could try a general solution of that form. Like Acos(wt) + Bsin(wt) and use that.

The book's solution was x=a+Acos(wt + B), so if A and a are constants of integration this should work. I'm not sure where B comes from though, or if I have to include the i in front of the sin in my general solution.

Yes, the general to the homogeneous solution is Acos(wt)+Bsin(wt). If you use the identity cos(a+b)=cos(a)cos(b)-sin(a)sin(b) on Acos(wt+B) you'll see that that form describes the same family of solutions.
 
  • #10
2,981
5
Yes, the general solution is Acos(wt)+Bsin(wt). If you use the identity cos(a+b)=cos(a)cos(b)-sin(a)sin(b) on Acos(wt+B) you'll see that that form describes the same family of solutions.

This is not true for the homogeneous equation posted in the op.

EDIT:

Never mind. :p
 

Related Threads on Second Order Differential Equation - Can't solve

Replies
7
Views
488
Replies
10
Views
7K
Replies
33
Views
1K
Replies
3
Views
2K
  • Last Post
Replies
1
Views
492
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
740
  • Last Post
Replies
3
Views
993
  • Last Post
Replies
1
Views
735
  • Last Post
Replies
3
Views
1K
Top