June 8th, 2019, 04:47 AM  #11 
Newbie Joined: Jun 2019 From: Moscow Posts: 14 Thanks: 0 
d/dx (2xyy^2/2+x^2/2)=x+2y d/dy (2xyy^2/2+x^2/2)=2xy int (x+2y)=x^2/2 + 2xy int (2xy)=x^2xy what's next? 
June 8th, 2019, 05:01 AM  #12 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
The second integration should have been with respect to y. A constant of integration is needed in the integrals. The constant can be a function of y when integrating with respect to x and vice versa. Edit: I should have said that the first integration should have been with respect to y. Last edited by skipjack; June 8th, 2019 at 10:34 AM. 
June 8th, 2019, 05:05 AM  #13  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,194 Thanks: 897 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \int (x + 2y)~dx = \dfrac{1}{2}x^2 + 2xy + F(y)$ where, so far, F(y) is arbitrary. You need to do the same thing with the second integral. $\displaystyle \int (2x  y)~dy = 2xy  \dfrac{1}{2}y^2 + G(x)$ Now you need to find possible F(y) and G(x) functions. Compare the two integrals. Dan  
June 8th, 2019, 05:29 AM  #14 
Newbie Joined: Jun 2019 From: Moscow Posts: 14 Thanks: 0 
how to find possible functions F (y) and G (x)?

June 8th, 2019, 07:09 AM  #15 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
Choose them so as to make the two integrals the same.

June 8th, 2019, 07:22 AM  #16 
Newbie Joined: Jun 2019 From: Moscow Posts: 14 Thanks: 0 
f(y) = y^2/2 g(x) = x^2/2 right? Last edited by skipjack; June 8th, 2019 at 11:21 AM. 
June 8th, 2019, 11:19 AM  #17 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
I referred to the wrong equation previously. You found ∂v/∂x = x + 2y and ∂v/∂y = 2x  y. As ∂u/∂y = ∂v/∂x, x + 2y needs now to be integrated with respect to y to give u. Thus u(x, y) = xy + y² + function of x. You correctly found u(x, y) = x²  xy + G(y) (where I've appended the G(y) term). For these two equations to be equivalent, you can choose G(y) = y² and "function of x" = x². Those choices give u(x, y) = x²  y²  xy, which is almost what you want. It's also necessary to supply an integer constant, namely 2, so that u(0, 0) = 2. Hence the required function of z is u + iv = x²  y²  xy + 2 + i(x²/2  y²/2 + 2xy), which can be written in terms of x + iy as (1 + (1/2)i)(x + iy)² + 2, i.e. (1 + i/2)z² + 2. 

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