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Separation of variables.

  • Thread starter hhhmortal
  • Start date
  • #1
176
0

Homework Statement



Hi, I dont really understand separation of variables very well, and I was hoping to do get my mind more clear on the following question:

(Q) Use separation of variables to find all the separable solutions of the equation:

d²y/dt² -c²(d²y/dx²) + w²y = 0

where 'w' and c are constants.




The Attempt at a Solution



I first started by saying the solution to this equation can be written as:

y = X(x)T(t)

so: X(d²T/dt²) - c²(d²X/dx²)T + w²XT

divide through by XT we get:

(1/T)d²T/dt² -(c²/X)(d²X/dx²) + w² = 0

What would be the next step? How can I find a solution from here?
 

Answers and Replies

  • #2
1,482
3
now you just have two independent DEQs to solve

(c²/X)(d²X/dx²) = 0

(1/T)d²T/dt² + w² = 0
 
  • #3
176
0
now you just have two independent DEQs to solve

(c²/X)(d²X/dx²) = 0

(1/T)d²T/dt² + w² = 0

Shouldn't it be -(c²/X)(d²X/dx²) = 0 ?

and shouldn't it be equated to a constant?

How would I go around to get a solution for these equations anyways?
 
  • #4
1,482
3
yea never mind, you should equate them to some constant

(c²/X)(d²X/dx²) = k

(1/T)d²T/dt² + w² = k
 
  • #5
Cyosis
Homework Helper
1,495
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Put one variable on one side and the other variable on the other side. For them to be equal they both need to be equal to the same constant! It is not some kind of lame trick to just equate them to a constant. Convince yourself that they must be equal to a constant.
 
  • #6
176
0
Put one variable on one side and the other variable on the other side. For them to be equal they both need to be equal to the same constant! It is not some kind of lame trick to just equate them to a constant. Convince yourself that they must be equal to a constant.
Oh ok. So now that I have both equated to the same constant, Do I just integrate both to get a solution or is there a general solution I can deduce from this?
 

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