How to Solve the Partial Differential Equation x(δu/δx)-(1/2)y(δu/δy)=0?

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Discussion Overview

The discussion revolves around solving the partial differential equation given by x(δu/δx) - (1/2)y(δu/δy) = 0. Participants explore methods for finding a separable solution and applying boundary conditions, specifically u(1,y) = 1 + sin(y). The conversation includes attempts to derive general solutions and fit them to the boundary condition.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant suggests looking for a separable solution of the form u(x, y) = X(x)Y(y) and expresses uncertainty about how to proceed with the solution.
  • Another participant proposes writing sin(y) as a power series to assist in finding the solution.
  • It is noted that each choice of a constant c yields a solution to the differential equation, and that linear combinations of such solutions also form solutions.
  • Several participants inquire about how to transition from the proposed form u(x, y) = A(x^c)(y^2c) to a general solution.
  • A series solution is presented, u(x,y) = ∑A_n x^{c_n} y^{2c_n}, with a follow-up question on how to match this to the boundary condition u(1,y) = ∑A_n y^{2c_n}.
  • Repeated requests for complete working solutions are made, indicating a desire for detailed guidance on the problem.
  • One participant expresses frustration at another's perceived lack of effort in attempting the problem, leading to a contentious exchange.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to solve the equation or how to apply the boundary condition. There are varying levels of understanding and willingness to engage with the problem, leading to disagreements about effort and capability.

Contextual Notes

Some participants express confusion about the application of power series and the steps needed to solve the differential equation, indicating potential gaps in foundational knowledge or assumptions about prior understanding.

Who May Find This Useful

This discussion may be useful for students or individuals interested in methods for solving partial differential equations, particularly those involving boundary conditions and series solutions.

EDerkatch
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x(δu/δx)-(1/2)y(δu/δy)=0

By first looking for a separable solution of the form u(x, y)=X(x)Y(y), find the general solution of the equation given above.

Determine the u(x,y) which satisfies the boundary condition u(1,y)=1+siny

For the separable form I have u(x, y)=A(x^c)(y^2c), could someone please show me how to do the rest of it.

Thank you.
 
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Write sin(y) as a power series.
 
Each choice of c gives you a solution of your diff.eq.
Since your diff.eq is linear, a sum of such solutions is also a solution of your diff.eq.
 
How do I get from u(x, y)=A(x^c)(y^2c) to the general solution?

Thank you.
 
All right:
A series solution of your diff.eq is:
[tex]u(x,y)=\sum_{n=1}^{\infty}A_{n}x^{c_{n}}y^{2c_{n}}[/tex],
whereby follows:
[tex]u(1,y)=\sum_{n=1}^{\infty}A_{n}y^{2c_{n}}[/tex]
and [itex]A_{n},c_{n}[/itex] are constants.

Now, how can you fit this expression for u(1,y) to the given boundary condition?
 
arildno said:
All right:
A series solution of your diff.eq is:
[tex]u(x,y)=\sum_{n=1}^{\infty}A_{n}x^{c_{n}}y^{2c_{n}}[/tex],
whereby follows:
[tex]u(1,y)=\sum_{n=1}^{\infty}A_{n}y^{2c_{n}}[/tex]
and [itex]A_{n},c_{n}[/itex] are constants.

Now, how can you fit this expression for u(1,y) to the given boundary condition?

Please go on...
 
Well, use my first hint in post 2.
 
arildno said:
Well, use my first hint in post 2.

Ok if you could please show me the COMPLETE working I would really appreciate it... Thank you.
 
arildno said:
Well, use my first hint in post 2.

Ok if you could please show me the COMPLETE working I would really appreciate it... Thank you.
 
  • #10
Do you know what a power series is? :confused:
 
  • #11
arildno said:
Do you know what a power series is? :confused:

Yes lol, I just can't do this question, could you please show me the working for it... In fact can you do it yourself?
 
  • #12
If you are not capable of doing basic algebra, you should not be attempting partial differential equations!

(Yes, I can do it myself! That's not really the point is it? You have been told exactly HOW to solve your equation, yet you have not even TRIED to apply what you have been told.)
 
  • #13
EDerkatch said:
Yes lol, I just can't do this question, could you please show me the working for it... In fact can you do it yourself?

That's it. I'm out of here. It is long since I've met a more ungrateful and lazy f*ckhead on PF as you.
 

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