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Partial Diff qn

  • Thread starter fredrick08
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  • #1
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Homework Statement


Does the following differential equation for u(x,y) have solutions which have the form of a product of functions of each independent variable:

[tex]\partial[/tex]2u/[tex]\partial[/tex]x[tex]\partial[/tex]y=u


3. The Attempt at a Solution [/]

=>[tex]\int[/tex][tex]\int[/tex]u dx dy =uxy=f(x)f(y) ???
 
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Answers and Replies

  • #2
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im not sure if that even makes sense, but if u(x,y)=f(x)f(y) and d*f(x)/dx=u/constant and d*f(y)/dy=constant... i think, i don't know im confused..
 
  • #3
Dick
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Set u(x,y)=f(x)*g(y). What is your partial derivative in terms of f(x), f'(x), g(y), and g'(y)?
 
  • #4
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so [tex]\partial[/tex]2u/[tex]\partial[/tex]x[tex]\partial[/tex]y=f(x)g(y)??
so in terms of f'(x) and g'(y) im not sure..
 
  • #5
Cyosis
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You have an equation that equates the derivative of u to u itself. You can already guess what u will be, but lets not guess.

Setting u(x,y)=f(x)g(y), then the equation becomes [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}=f(x)g(y)[/tex]. Now what Dick asks you to do is to calculate [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}[/tex].

Hint:

[tex]
\frac{d}{dx}f(x)= ?
[/tex]

In terms of f'(x)
 
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  • #6
Dick
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I actually meant u(x,y)=f(x)g(y). There's no reason why the function of x would have to be the same as the function of y. And yes, evaluate the derivative.
 
  • #7
Cyosis
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Sorry, that was sloppy of me I actually meant f(x)g(x). I will fix it.
 
  • #8
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hmm d/dx of f(x).. isn't that just f'(x)??.... is it something like the integral of f'(x)g'(y)??
 
  • #9
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so... [tex]
\frac{\partial^2 f(x)g(y)}{\partial x \partial y}
[/tex] =[tex]\int[/tex][tex]\int[/tex]f'(x)g'(y) dx dy??
 
  • #10
Cyosis
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What you've written down now is the equivalent to [itex]\frac{d}{dx}f(x)=\int f'(x) dx[/itex]. Which is wrong. Lets test it, f(x)=x then [itex]\frac{d}{dx}x=1 \neq \int 1 dx=x+c[/itex].

hmm d/dx of f(x).. isn't that just f'(x)??
This is right, so if you evaluate [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}[/tex] you get....?
 
  • #11
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u get f'(x)g'(y)?? Im confused becasue when u take partial of f(x), u get d/dy(d/dx*f(x))=0??
 
  • #12
Cyosis
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Yes that's correct, but you're not taking the partial of f(x), you're talking the partial derivative of f(x)g(y).

[tex]
\frac{\partial^2 f(x)g(y)} {\partial x \partial y}=\frac{\partial}{\partial x} f(x) \frac{\partial}{\partial y}g(y)=f'(x)g'(y)
[/tex]
 
  • #13
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ok so since f(x)g(y)=f'(x)g(y)=u(x,y)=??
 
  • #14
Cyosis
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You forgot a prime it should be [itex]f'(x)g'(y)=f(x)g(y)=u(x,y)[/itex]. You can instantly see which function satisfies the equation, but you can also separate the variables.
 
  • #15
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e^x+y?? so f(x)=e^x and g(y)=e^y same as the derivativeS?
 
  • #16
Cyosis
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That would be a solution indeed, you're on the right track. it is not the general solution however. The general solution will involve integration constants and a constant in the argument of the exponential. Can you see how to integrate the differential equation?
 
  • #17
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e^xy?? so f(x)=e^x and g(y)=e^y same as the derrivativeS?

srry this was an accident
 
  • #18
Cyosis
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Don't randomly guess, [itex]f'(x)g'(y)=y e^{xy}xe^{xy} \neq e^{xy}[/itex]. Separate the variables first, that is all the x dependent stuff on one side and all the y dependent stuff on the other side.
 
  • #19
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na im confused now... like f(x)=e^c1 g(y)=e^c2 and u(x,y)=e^c1+c2??
 
  • #20
Cyosis
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If c are constants then f'(x)=g'(y)=0, which doesn't fit either. Can you please separate the variables first.
 
  • #21
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im confused because when u take the derivative of a constant its zero?
 
  • #22
Cyosis
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Yes when you take the derivative of a constant it's zero. You may want to elaborate on your confusion.
 
  • #23
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oh ok i will try we only just learned this... U(x,y)=F(x)G(y)

F'(x)G'(y)=F(x)G(y) divide by U(x,y)=F'(x)G'(y)/F(x)G(y)=1
im not sure the ones i have done arent as confusing as this one
 
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  • #24
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no thats not rite... umm F'/F=G/G'=gamma?
 
  • #25
Cyosis
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No you have the y and x terms on the same side now. Separation of variables means that you put all the functions depending on x on one side and all functions depending on y on the other side. Can you do this?
 

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