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October 4th, 2019, 09:07 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  PDEs
Solve the equations below : (a) $\displaystyle \partial z(x,y) =x^{1} +y^{1} $. (b) $\displaystyle \partial_{x} \partial _{y} Z(x,y)=xy+x+1$. Last edited by idontknow; October 4th, 2019 at 09:18 AM. 
October 4th, 2019, 10:48 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 
Note :$\displaystyle \partial_{x}$ is the partial derivative and $\displaystyle \partial z =z_{x} ' +z_{y} '$.(trying to not use complicated symbols ) .
Last edited by idontknow; October 4th, 2019 at 10:51 AM. 
October 4th, 2019, 11:00 AM  #3 
Senior Member Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 
Well, I was totally confused, and still have my doubts. Are you saying the following: $\displaystyle \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = \frac 1x + \frac 1y$ $\displaystyle \frac{\partial Z^2}{\partial x \partial y} = xy+x+1$ ...where $z$ and $Z$ are functions of (x,y)? 
October 4th, 2019, 12:47 PM  #4  
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  Quote:
 
October 4th, 2019, 12:56 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,588 Thanks: 1432 
$z_x + z_y = \dfrac 1 x + \dfrac 1 y$ let $z(x,y) = \log(x) + \log(y) + h(x,y),~\ni h_x+h_y = 0$ $z_x + z_y = \dfrac 1 x + \dfrac 1 y$ $h(x,y) = c(xy)$ satisfies $h_x + h_y=0$ $z(x,y) = \log(x) + \log(y) + c(xy)$ Last edited by romsek; October 4th, 2019 at 01:02 PM. 
October 4th, 2019, 01:01 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,588 Thanks: 1432 
$Z_{yx}(x,y) = xy + x + 1 = x(y+1) + 1$ Integrate over $x$ $Z_y(x,y) = \dfrac{y+1}{2}x^2 + x + c,~c \in \mathbb{R}$ Integrate over $y$ $Z(x,y) =\dfrac 1 4 (y+1)^2 x^2 + (x+c) y + d,~c,d \in \mathbb{R}$ 

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