Multi-variable PDEs

Dec 2015
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166
Earth
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:
Dec 2015
1,076
166
Earth
Note :\(\displaystyle \partial_{x}\) is the partial derivative and \(\displaystyle \partial z =z_{x} ' +z_{y} '\).(trying to not use complicated symbols ) .
 
Last edited:
Jun 2019
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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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Dec 2015
1,076
166
Earth
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.
 

romsek

Math Team
Sep 2015
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1,673
USA
$z_x + z_y = \dfrac 1 x + \dfrac 1 y$

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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romsek

Math Team
Sep 2015
2,958
1,673
USA
$Z_{yx}(x,y) = xy + x + 1 = x(y+1) + 1$

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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