Finding Separable Solutions for a Second-Order Partial Differential Equation

In summary, The conversation is about using separation of variables to find all the separable solutions of a given equation. The solution can be written as y = X(x)T(t), which leads to two independent differential equations. To solve these equations, one variable is put on one side and the other variable is put on the other side, equating them both to a constant. This constant can then be integrated to find the general solution.
  • #1
hhhmortal
176
0

Homework Statement



Hi, I don't 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?
 
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  • #2
now you just have two independent DEQs to solve

(c²/X)(d²X/dx²) = 0

(1/T)d²T/dt² + w² = 0
 
  • #3
waht said:
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
yea never mind, you should equate them to some constant

(c²/X)(d²X/dx²) = k

(1/T)d²T/dt² + w² = k
 
  • #5
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
Cyosis said:
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?
 

Related to Finding Separable Solutions for a Second-Order Partial Differential Equation

1. What is separation of variables?

Separation of variables is a mathematical technique used to solve certain types of partial differential equations. It involves breaking down a complex equation into simpler equations that each have only one variable.

2. Why is separation of variables important?

Separation of variables is important because it allows us to solve otherwise unsolvable partial differential equations. It is also a fundamental concept in many areas of physics and engineering, such as fluid dynamics and heat transfer.

3. How does separation of variables work?

To use separation of variables, we first assume that the solution to the partial differential equation can be written as a product of functions, each of which depends on only one variable. We then substitute this form into the original equation and rearrange terms until we have separate equations for each variable.

4. What types of problems can be solved using separation of variables?

Separation of variables is typically used to solve boundary value problems, which involve finding a solution that satisfies certain conditions at the boundaries of a physical system. It is also used to solve initial value problems, where the solution is determined by specifying the initial conditions at a particular point in time or space.

5. Are there any limitations to using separation of variables?

While separation of variables is a powerful tool for solving certain types of partial differential equations, it is not applicable to all equations. In some cases, the assumptions made to separate the variables may not hold, or the resulting equations may be too difficult to solve. Other methods, such as numerical techniques, may be needed in these situations.

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