Multi-variable PDEs

idontknow

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:

idontknow

Note :$$\displaystyle \partial_{x}$$ is the partial derivative and $$\displaystyle \partial z =z_{x} ' +z_{y} '$$.(trying to not use complicated symbols ) .

Last edited:

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)?

1 person

idontknow

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
$z_x + z_y = \dfrac 1 x + \dfrac 1 y$

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 $h_x + h_y=0$

$z(x,y) = \log(x) + \log(y) + c(x-y)$

Last edited:
1 person

romsek

Math Team
$Z_{yx}(x,y) = xy + x + 1 = x(y+1) + 1$

Integrate over $x$

$Z_y(x,y) = \dfrac{y+1}{2}x^2 + x + c,~c \in \mathbb{R}$

Integrate over $y$

$Z(x,y) =\dfrac 1 4 (y+1)^2 x^2 + (x+c) y + d,~c,d \in \mathbb{R}$

1 person