The Student Room Group
Reply 1
find d/dx and d/dy in terms of d/du and d/dv by using the chain rule. Then use them in the equation in finding the second derivatives and you should find alot will cancel and you have some easy expression to intergrate up. (well done in the cup by the way)

You should get
d/dx = (d/du + d/dv)(1/2)
d/dy = (d/du - d/dv)(1/2)

in the equation this should give you
(1/4)(d/du + d/dv)(df/du + df/dv) - (1/4)(d/du - d/dv)(df/du - df/dv) = u+v
d2f/dudv=u+v

intergrate up
df/du = uv + 1/2(v^2) + f(u)
f = 1/2(vu^2) + 1/2(uv^2) + F(u) + G(v)
f = (1/2)uv(u+v) + F(U) + G(v)
where F and G are arbitary funtions

you can then give this in terms of x and y
Reply 2
Retropmot

You should get
d/dx = d/du + d/dv
d/dy = d/du - d/dv


thanks

i did get up to that and then got a bit lost :redface:

but now i understand it

many thanks :yy: (rep on its way tomorrow)
Reply 3
davemarkey
thanks

i did get up to that and then got a bit lost :redface:

but now i understand it

many thanks :yy: (rep on its way tomorrow)

actually i made a mistake check my post again du/dx and dv/dx aren't what i said
u = (x + y)/2 so du/dx = 1/2
v = (x - y)/2 so dv/dx = 1/2
Reply 4
thanks

i am doing it myself now anyways as i know what to do :smile:
Reply 5
Retropmot
(well done in the cup by the way)


:wink:

cheers

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