higher order partial derivatives


by magnifik
Tags: derivatives, order, partial
magnifik
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#1
Jun8-10, 01:19 PM
P: 361
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
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tiny-tim
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#2
Jun8-10, 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, 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 ).
magnifik
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#3
Jun8-10, 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!

magnifik
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#4
Jun8-10, 01:57 PM
P: 361

higher order partial derivatives


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?
HallsofIvy
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#5
Jun8-10, 01:57 PM
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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 [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].
magnifik
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#6
Jun8-10, 02:01 PM
P: 361
Quote Quote by HallsofIvy View Post
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 [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].

THANK YOU!! makes so much sense now :)
tiny-tim
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#7
Jun8-10, 02:03 PM
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Quote Quote by magnifik View Post
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?
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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