Question

Eliminate the arbitrary constants indicated in brackets from the following equation and form corresponding partial differential equation :

\frac{1}{e^{z-(\frac{x^2}{y})}}= \frac{ax^2}{y^2} + \frac{b}{y}

Answer 1

$\displaystyle \frac{1}{e^{z-(\frac{x^2}{y})}}= \frac{ax^2}{y^2} + \frac{b}{y}$ (1)

Differentiating (1) by x:

$\displaystyle e^{-z + \frac{x^2}{y}} (\frac{ \partial z}{\partial x}+\frac{2x}{y}) = \frac{2ax}{y^2}$ , from here

$\displaystyle a = e^{-z + \frac{x^2}{y}} \frac{y^2}{2x}(\frac{ \partial z}{\partial x}+\frac{2x}{y})$

Differentiating (2) by y:

$\displaystyle e^{-z + \frac{x^2}{y}} (-\frac{ \partial z}{\partial y}-\frac{x^2}{y^2}) = -\frac{2ax^2}{y^3} - \frac{b}{y^2}$

$b = \displaystyle e^{-z + \frac{x^2}{y}} y^2 (\frac{ \partial z}{\partial y}+\frac{x^2}{y^2}) -\frac{2ax^2}{y} = e^{-z + \frac{x^2}{y}} [y^2(\frac{ \partial z}{\partial y}+\frac{x^2}{y^2}) -xy(\frac{ \partial z}{\partial x}+\frac{2x}{y})]$

Then (1) transforms to

$\displaystyle e^{-z + \frac{x^2}{y}} = e^{-z + \frac{x^2}{y}} \frac{y^2 \cdot x^2}{2x \cdot y^2}(\frac{ \partial z}{\partial x}+\frac{2x}{y}) + e^{-z + \frac{x^2}{y}} [y(\frac{ \partial z}{\partial y}+\frac{x^2}{y^2}) -x(\frac{ \partial z}{\partial x}+\frac{2x}{y})]$

$\displaystyle 1 = \frac{ x}{2}(\frac{ \partial z}{\partial x}+\frac{2x}{y}) + [y(\frac{ \partial z}{\partial y}+\frac{x^2}{y^2}) -x(\frac{ \partial z}{\partial x}+\frac{2x}{y})]$

$\displaystyle 1 = \frac{ x}{2}(\frac{ \partial z}{\partial x})+\frac{x^2}{y} + y(\frac{ \partial z}{\partial y})+\frac{x^2}{y} -xy(\frac{ \partial z}{\partial x}) - \frac{2x^2}{y}$

$\displaystyle 1 = (\frac{ x}{2}-xy)(\frac{ \partial z}{\partial x})+ y(\frac{ \partial z}{\partial y})$

$\displaystyle (\frac{ x}{2}-xy)(\frac{ \partial z}{\partial x})+ y(\frac{ \partial z}{\partial y}) =1$