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Trying to get this PDE in terms of 'y'by jaketodd
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#19
Jun2411, 04:03 AM

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So if I asked you to find y when:
[tex] \frac{dy}{dx}+2xy=x [/tex] Could you tell me how to calculate y? Solving PDEs such as this is straightforward, if you have the right experience. It looks to me as if you're jumping way ahead of your experience. I would humbly suggest that you start with how to solve the equation I gave you before you start on the wave equation. On an aside, I would also suggest you look how to solve 1st order partial differential equations before you look into 2nd order PDEs. 


#20
Jun2411, 08:21 AM

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#21
Jun2411, 08:25 AM

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My advice would be to get a book on differential equations, and start reading it. If you get stuck on some concepts then by all means ask and we will help you.



#22
Jun2411, 11:23 AM

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I wish someone could just show me the steps involved... I'll transfer $50 with Paypal to the first person who does this for me; anyone who provides the steps of solving for y from the original post with all variables definable (for example, no unsolvable constants). It's for a physics paper I'm working on; I don't need to understand all the mathematical concepts, I just need the equation solved for y please.
Jake 


#23
Jun2411, 07:22 PM

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Please help guys; it's for a physics paper and I really don't need to understand all the math.



#24
Jun2411, 08:02 PM

P: 737

The arbitrary constants are arbitrary. That means the equation is true no matter their value. Think about what taking a derivative does to a constant, it's like that.
For example, if y'=k, where k is constant, then y=kx+C, where C is an arbitrary constant. As you can see, no matter what C is, since it is constant, it just "goes away," and the derivative of kx+C is k, which is what we wanted. So, what I'm saying is, the arbitrary constants are just that. You can't solve for them, because they're true for all real numbers ie, they're arbitrary. 


#25
Jun2511, 02:01 AM

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I see what Tyler means. So those equations in earlier posts  are they enough to solve for y? Could someone put it all together for me? $50 on the line.



#26
Jun2511, 11:11 PM

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A lot of thought went into this. I don't know if it's right, but it's the best I can do right now...
Do I put these: Into these: How do the C's go away? And how do I solve for k? Thanks, Jake 


#27
Jun2511, 11:29 PM

P: 737

Yes. It should be e^{x}
No. The equations you quoted from me are the solutions to those you quoted from hunt_mat. To get y, you multiply X(x) and T(t), like I did in the third equation in my post you quoted. The C's don't go away. They're part of the answer. You CAN'T get the answer in terms of ONLY the original variables. They HAVE to be there. 


#28
Jun2611, 12:43 AM

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Is there a way of solving for the C's? I can define all the variables when y = 0. However, would that make all the C's zero if there is a time, t, when y = 0? Thanks, Jake 


#29
Jun2611, 04:15 AM

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The two solutions of the differential equations should be:
[tex] X(x)=Ae^{x\sqrt{k}}+Be^{x\sqrt{k}},\quad T(t)=Ce^{\nu t\sqrt{k}}+De^{\nu t\sqrt{k}} [/tex] So the solution for y becomes: [tex] y=(Ae^{x\sqrt{k}}+Be^{x\sqrt{k}})(Ce^{\nu t\sqrt{k}}+De^{\nu t\sqrt{k}}) [/tex] But there are two things you need to consider, the sign of k, as this will decide what sort of solutions you will get (do you want oscillatory solutions ([itex]k=\lambda^{2}[/itex]), then there is the fact that the equation is linear and you can just add solutions together to find a new solution, so if y_1 and y_2 solve your equation then so does y_3=y_1+y_2 (exercise). So the next thing you require is initial and boundary conditions. 


#30
Jun2611, 11:26 PM

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So for oscillatory solutions with k=lambda^2, wouldn't that produce imaginary numbers?
Thanks, Jake 


#31
Jun2711, 04:08 AM

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Go back to the original equations:
[tex] X''(x)+\lambda^{2}X(x)=0,\quad T''(t)+\lambda^{2}T(t)=0 [/tex] What are the solutions to these equations? 


#32
Jun2711, 06:04 AM

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An n^{th} order ordinary differential equation always involves n unknown constants. An n^{th} order partial differential equation always involves n unknown functions.
I pretty much told you before that the general solution to that differential equation is y= F(x vt)+ G(x+ vt) where F and G are arbitrary, twice differentiable, functions. The simple fact is that you are trying to do a problem that is far beyond your knowledge. If you do not know how to solve basic ordinary differential equations, no one is going to be able to tell you how to solve partial differential equations in a few sentences. 


#33
Jun2711, 08:57 AM

P: 343

I agree with Ivy. Everyone here is trying to help you (the OP) solve the problem, but it's your inexperience that's stopping you from understanding the solution process. PDEs are fairly difficult to solve a lot of the time even for people with a fair amount of math under their belt. I do appreciate your bravado in tackling the wave equation and applaud your desire to extend your mathematical prowess, however I think you should spend a bit more time learning the fundamentals.



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