Question

from the partial differential equation by eliminating the arbitary constant from the equation 1. z=ax + by+ cxy​

Answer 1

Basic notations :-

$\boxed{ \bf \: \dfrac{ \partial \: z}{\partial \: x} = p}$

$\boxed{ \bf \: \dfrac{ \partial \: z}{\partial \: y} = q}$

$\boxed{ \bf \: \dfrac{ \partial^{2} \: z}{\partial \: x^{2} } = r}$

$\boxed{ \bf \: \dfrac{ \partial^{2} \: z}{\partial \: y \: \partial \: x} =\dfrac{ \partial^{2} \: z}{\partial \: x \: \partial \: y} = s}$

$\boxed{ \bf \: \dfrac{ \partial^{2} \: z}{\partial \: y^{2} } = t}$

Let's solve the problem now!!

Given that

$\bf :\longmapsto\:z \: = ax + by + cxy - - - (1)$

On differentiating partially w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{\partial \: z}{\partial \: x} = a + cy - - (2)$

On differentiating partially w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{ {\partial}^{2}z }{ {\partial \: x}^{2}} = 0 \implies\: r \: = 0 - - - (3)$

$\bf\implies \:r \: = \: 0 \: is \: the \: required \: differential \: equation.$

Again,

On differentiating partially (1) w. r. t. y, we get

$\rm :\longmapsto\:\dfrac{\partial \: z}{\partial \: y} = b + cx - - (4)$

On differentiating partially w. r. t. y, we get

$\rm :\longmapsto\:\dfrac{ {\partial}^{2}z }{ {\partial \: y}^{2}} = 0 \implies \: t = 0 - - - (5)$

$\bf\implies \:t \: = \: 0 \: is \: the \: required \: differential \: equation.$

Again,

On differentiating partially equation (2) w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{ {\partial}^{2}z }{ \partial \: y{\partial \: }x} = c - - (6)$

Multiply equation (2) by x and equation (4) by y and add,

$\rm :\longmapsto\:x\dfrac{\partial \: z}{\partial \: x} + y\dfrac{\partial \: z}{\partial \: y} = ax + cxy + by + cxy$

$\rm :\longmapsto\:xp + yq = z + xy\dfrac{ {\partial}^{2}z }{ \partial \: y{\partial \: }x}$

$\bf\implies \:xp + yq = z + xys \: is \: differential \: equation.$

Thus, we have 3 differential equations of order 2, represented by equations as follow :-

$\rm :\longmapsto\:r = 0$

$\rm :\longmapsto\:t = 0$

$\rm :\longmapsto\:px + qy = z + xys$

Remark :-

From the above calculations, we observed that if the number of arbitrary constants to be eliminated from the given is more than the number of independent variables, then there exists partial differential equations of order 2 or of higher order.