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October 4th, 2019, 09:07 AM   #1
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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.
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October 4th, 2019, 10:48 AM   #2
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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.
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October 4th, 2019, 11:00 AM   #3
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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)?
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October 4th, 2019, 12:47 PM   #4
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Quote:
Originally Posted by DarnItJimImAnEngineer View Post
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.
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October 4th, 2019, 12:56 PM   #5
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$z_x + z_y = \dfrac 1 x + \dfrac 1 y$

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 $h_x + h_y=0$

$z(x,y) = \log(x) + \log(y) + c(x-y)$
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Last edited by romsek; October 4th, 2019 at 01:02 PM.
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October 4th, 2019, 01:01 PM   #6
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$Z_{yx}(x,y) = xy + x + 1 = x(y+1) + 1$

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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