Partial Diff Qn: Can u(x,y) be a Product of fx and fy?

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The discussion revolves around solving the partial differential equation \(\partial^2 u/\partial x \partial y = u\) by assuming a solution of the form \(u(x,y) = f(x)g(y)\). Participants explore the implications of this assumption, leading to the conclusion that the functions \(f(x)\) and \(g(y)\) must satisfy the relationship \(\frac{f'(x)}{f(x)} = \frac{g'(y)}{g(y)} = \gamma\), where \(\gamma\) is a constant. The general solution is derived as \(u(x,y) = Ae^{\gamma x}Be^{\gamma y}\), confirming that the product form is indeed valid for this differential equation.

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  • #61
can i just flip it?? then it will be the same as f... =ln(g(y))?
 
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  • #62
Yes you can flip one side, but then you also need to flip the other side.
 
  • #63
ok so ln(g(y))=(ax+c)^-1??
 
  • #64
No you are solving the differential equation g(y)/g'(y)=a , then g'(y)/g(y)=?
 
  • #65
1/a? so ln(g(y))=(y/a)+c?? =>g(y)=e^((y/a)+c)
 
  • #66
No. How can the integral of f'/f have a different value than g'/g. If g(y)/g'(y)=a then g'(y)/g(y)=1/a. Now integrate.
 
  • #67
yes isn't that what i did? int(1/a)dy=(y/a)+c right?
 
  • #68
Ugh sorry it's 4 am here I am getting sleepy! Yes you're right so what is the the total solution to the partial differential equation you began with?
 
  • #69
u(x,y)=f(x)g(y)=Ae^(ax)*Be^(y/a)... but I am not sure... if i differentiate that, i don't thik it works out.
 
  • #70
It's correct, just differentiate it and you will see it will turn out correctly.
 
  • #71
ok omg thankyou very much... very much appreciated = ) have a good sleep
 
  • #72
You're welcome, thanks and good night!
 

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