Question

Eliminate the arbitrary constants indicated in brackets from the following equation

and form corresponding partial differential equation:

e

1

{z−(

x2

y

)}

=

ax

2

y2 +

b

y

;(a,b)

Answer 1

$\displaystyle e^{z - x^2y} = ax^2y^2 + by\\ e^{z - x^2y}\left(\frac{\partial z}{\partial x} - 2xy\right) = 2axy^2\\ e^{z - x^2y}\left(\frac{\partial z}{\partial y} - x^2\right) = 2ax^2y + b\\ e^{z - x^2y}\left(\left(\frac{\partial z}{\partial x} - 2xy\right)^2 + \frac{\partial^2 z}{\partial x^2} - 2x\right) = 2ay^2\\ e^{z - x^2y}\left(\left(\frac{\partial z}{\partial y} - x^2\right)^2 + \frac{\partial^2 z}{\partial y^2}\right) = 2ax^2\\ \frac{e^{z - x^2y}}{y^2}\left(\left(\frac{\partial z}{\partial x} - 2xy\right)^2 + \frac{\partial^2 z}{\partial x^2} - 2x\right) = \frac{e^{z - x^2y}}{x^2}\left(\left(\frac{\partial z}{\partial y} - x^2\right)^2 + \frac{\partial^2 z}{\partial y^2}\right)\\ x^2\left(\frac{\partial z}{\partial x} - 2xy\right)^2 + x^2\frac{\partial^2 z}{\partial x^2} - 2x^3 = y^2\left(\frac{\partial z}{\partial y} - x^2\right)^2 + y^2\frac{\partial^2 z}{\partial y^2}\\ \left(x\frac{\partial z}{\partial x} - 2x^2y - y\frac{\partial z}{\partial y} + x^2y\right)\left(x\frac{\partial z}{\partial x} - 2x^2y + y\frac{\partial z}{\partial y} - x^2y\right) + x^2\frac{\partial^2 z}{\partial x^2} - y^2\frac{\partial^2 z}{\partial y^2} - 2x^3 = 0$