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Higher order partial derivatives 
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#1
Jun810, 01:19 PM

P: 361

Consider the partial dierential equation, (y^{4}x^{2})u_{xx}  2xyu_{xy}  y^{2}u_{yy} = 1. We will make the substitution x = s^{2}  t^{2} 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)(st) = 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
Jun810, 01:47 PM

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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, st = y, so 2s = x/y + y = (x + y^{2})/y, 2t = (x  y^{2})/y. To get the solution, you need to find u_{ss} u_{st} and u_{tt} (and I expect u_{st} will be zero, which will make the solution fairly easy ). 


#3
Jun810, 01:54 PM

P: 361

errr.. i meant the solution for part a, which was putting x and y in terms of s and t. thanks!



#4
Jun810, 01:57 PM

P: 361

Higher order partial derivatives
but i don't understand how you got from s+t = x/y, st = y to 2s and 2t...how did you combine the previous two equations to get that?



#5
Jun810, 01:57 PM

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Excuse me, tinytim, 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 [itex]x= s^2 t^2[/itex] and [itex]y= s t[/itex]. From [itex]y= s t[/itex] you can say that [itex]s= y+ t[/itex] and, putting that into [itex]x=s^2 t^2[/itex], [itex]x= y^2+2yt+ t^2 t^2= y^2+ 2yt[/itex]. Solve that equation for t, then replace t by that expression in [itex]s= y+ t[/itex]. 


#6
Jun810, 02:01 PM

P: 361

THANK YOU!! makes so much sense now :) 


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