text: # Higher order partial derivatives
  
  1. Jun 8, 2010
  
  ### magnifik
  
  Consider the partial di erential equation, (y4-x2)uxx - 2xyuxy - y2uyy = 1. We will make the substitution x = s2 - t2 and y = s - t, to simplify
  
  (a) Solve for s and t as functions of x and y
  
  the farthest point i got to was
  x = s^2 - t^2 = (s+t)(s-t) = y(s+t)
  y = s - t
  s+t = x/y
  
  i don't know what to do after that.. i have the solution, but i have no idea how to get to it.
  the solution is
  s = x + y^2 / 2y
  t = x - y^2 / 2x
  
  2. Jun 8, 2010
  
  ### tiny-tim
  
  Hi magnifik!
  
  No, you're mssing the point …
  
  that isn't the solution, it's just writing s and t in terms of x and y …
  
  s+t = x/y, s-t = y, so 2s = x/y + y = (x + y2)/y, 2t = (x - y2)/y.
  
  To get the solution, you need to find uss ust and utt (and I expect ust will be zero, which will make the solution fairly easy ).
  
  3. Jun 8, 2010
  
  ### magnifik
  
  errr.. i meant the solution for part a, which was putting x and y in terms of s and t. thanks!
  
  4. Jun 8, 2010
  
  ### magnifik
  
  but i don't understand how you got from s+t = x/y, s-t = y to 2s and 2t...how did you combine the previous two equations to get that?
  
  5. Jun 8, 2010
  
  ### HallsofIvy
  
  Excuse me, tiny-tim, but the only question asked was "(a) Solve for s and t as functions of x and y".
  
  Presumably, there is a "b" but so far all we want to do is solve for s and t.
  
  magnifik, you have $x= s^2- t^2$ and $y= s- t$.
  
  From $y= s- t$ you can say that $s= y+ t$ and, putting that into $x=s^2- t^2$, $x= y^2+2yt+ t^2- t^2= y^2+ 2yt$. Solve that equation for t, then replace t by that expression in $s= y+ t$.
  
  6. Jun 8, 2010
  
  ### magnifik
  
  THANK YOU!! makes so much sense now :)
  
  7. Jun 8, 2010
  
  ### tiny-tim
  
  erm … I added, and subtracted!
  
  Add s+t to s-t, you get 2s; subtract, you get 2t …
  
  this is a common transformation, and you should be familiar with it.
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