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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 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)? Thanks from idontknow October 4th, 2019, 12:47 PM   #4
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 Originally Posted by DarnItJimImAnEngineer 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)?
Yes. 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 $z(x,y) = \log(x) + \log(y) + h(x,y),~\ni h_x+h_y = 0$ $z_x + z_y = \dfrac 1 x + \dfrac 1 y$ $h(x,y) = c(x-y)$ satisfies $h_x + h_y=0$ $z(x,y) = \log(x) + \log(y) + c(x-y)$ Thanks from idontknow 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 $y$ $Z(x,y) =\dfrac 1 4 (y+1)^2 x^2 + (x+c) y + d,~c,d \in \mathbb{R}$ Thanks from idontknow Tags multivariable, pdes 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 Chloesannon Real Analysis 5 June 30th, 2018 11:12 AM irunktm Algebra 6 September 7th, 2012 12:42 PM Chasej Calculus 7 September 16th, 2011 12:18 AM person1200 Calculus 1 September 12th, 2010 07:17 PM wetmelon Algebra 2 March 23rd, 2009 11:00 PM

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