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 June 8th, 2019, 04:47 AM #11 Newbie   Joined: Jun 2019 From: Moscow Posts: 14 Thanks: 0 d/dx (2xy-y^2/2+x^2/2)=x+2y d/dy (2xy-y^2/2+x^2/2)=2x-y int (x+2y)=x^2/2 + 2xy int (2x-y)=x^2-xy 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. Thanks from topsquark Last edited by skipjack; June 8th, 2019 at 10:34 AM.
June 8th, 2019, 05:05 AM   #13
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 Originally Posted by Lasqa d/dx (2xy-y^2/2+x^2/2)=x+2y d/dy (2xy-y^2/2+x^2/2)=2x-y int (x+2y)=x^2/2 + 2xy int (2x-y)=x^2-xy what's next?
Remember these are partial differential equations. So:
$\displaystyle \int (x + 2y)~dx = \dfrac{1}{2}x^2 + 2xy + F(y)$

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. Thanks from topsquark
 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. Thanks from topsquark and Lasqa

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