Question

y.δ²y/δxδy+2.δu/δx=0
w.r.t u(x,1)=x²+1, u(0,y)=e^y

Answer 1

I had my own work to do, lol.

I suppose the question's really asking to determine \(u(x,y)\). I'm thinking it through.

Answer 2

lol @Ishaan94 now it looks like I'm talking to myself :P

Answer 3

Sorry :/

Answer 4

I think I'll live ;D

Answer 5

Darn, this is either difficult, or I'm not reading the partials right. Whichever one it is, @TuringTest @FoolForMath @satellite73 @myininaya .

See? I'm a part of the problem solving process. :P

I get others to do it instead.

Answer 6

@ayaz110 Could you clarify your question? http://www.codecogs.com/latex/eqneditor.php

Answer 7

partial derivative with respect to --------

Answer 8

u(1,0) should get us 1,e

Answer 9

Ux(x,1) = f'(x,y) = x^2 + y

U(x,y) = 1/3 x^3+ yx + g(x,y)

Uy = g'(x,y) = x+e^y

Uy(0,y) = e^y

Answer 10

something like this i think, notations bad tho but it helped me focus :)

Answer 11

U(x,y) = 1/3 x^3 + xy + e^y

Answer 12

only thing to do is take the required derivatives and see if it fits

Answer 13

y.δ²y/δxδy+2.δu/δx=0

is that dy/dxdy a typo?

Answer 14

2.Ux = 2x^2 +2y

y Uyx = yx

those dont match up to zero tho

Answer 15

oh well, i think itll be someting similar to that

Answer 16

I ran into a similar problem, I think there's a typo?

Answer 17

I'm glad someone else ran into this problem; it confirms that I'm not just derping.