Question

The PDE formed by eliminating the arbitrary constants a and b from the
relation 2z = (x2/a2) + (y2/b2) is
(C04 – Comprehension)​

Answer 1

Answer:

arbitrary constants or by the elimination of arbitrary functions. Solution of ... Formation of partial differential equations by eliminating ... and b from z = (x2 + a2)(y2 + b2).

Answer 2

Given,

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

Solution,

Differentiate equation (1) with respect to x,

$\[\begin{array}{l}2z = \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}\\ \Rightarrow 2\frac{{dz}}{{dx}} = \frac{{2x}}{{{a^2}}}\\ \Rightarrow \frac{{{d^2}z}}{{d{x^2}}} = \frac{1}{{{a^2}}} - - - - \left( 2 \right)\end{array}\]$

Differentiate equation (1) with respect to y,

$\[\begin{array}{l}2z = \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}\\ \Rightarrow 2\frac{{dz}}{{dy}} = \frac{{2y}}{{{b^2}}}\\ \Rightarrow \frac{{{d^2}z}}{{d{y^2}}} = \frac{1}{{{b^2}}} - - - - \left( 3 \right)\end{array}\]$

Put equation 2 and 3 in equation 1.

$\[2z = {x^2}\frac{{{d^2}z}}{{d{y^2}}} + {y^2}\frac{{{d^2}z}}{{d{y^2}}}\]$

Hence the PDF is $\[2z = {x^2}\frac{{{d^2}z}}{{d{y^2}}} + {y^2}\frac{{{d^2}z}}{{d{y^2}}}\]$