# higher order partial derivatives

by magnifik
Tags: derivatives, order, partial
 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
 Sci Advisor HW Helper Thanks P: 26,167 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 ).
 P: 361 errr.. i meant the solution for part a, which was putting x and y in terms of s and t. thanks!
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?
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,898 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 $s= y+ t$.
P: 361
 Quote by 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 $s= y+ t$.

THANK YOU!! makes so much sense now :)