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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: 20,966 Thanks: 2216 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: 20,966 Thanks: 2216 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: 20,966 Thanks: 2216 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 Tags analyticity, restore Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ach4124 Complex Analysis 0 November 8th, 2015 02:47 AM chris2401 Complex Analysis 3 February 6th, 2013 12:40 PM hanahou Complex Analysis 1 October 28th, 2007 12:55 PM

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