To form a partial differential equation from a given relation, we need to eliminate the arbitrary constants present in that relation. In this case, the given relation is \(z = (x^2 + a)(y^2 + b)\), and the arbitrary constants are 'a' and 'b'. We will achieve this by calculating partial derivatives.

Derivatives Calculation: Finding Partial Derivatives

First, let's find the partial derivatives of \(z\) with respect to \(x\) and \(y\). We will denote \(\frac{\delta z}{\delta x}\) as \(p\) and \(\frac{\delta z}{\delta y}\) as \(q\).

• Partial derivative of z with respect to x (\(\frac{\delta z}{\delta x}\) or \(p\)):

  When differentiating with respect to \(x\), we treat \(y\) and the constant \(b\) as constants.

  \[ p = \frac{\delta z}{\delta x} = \frac{\delta}{\delta x} [(x^2 + a)(y^2 + b)] \]

  Since \((y^2 + b)\) is a constant with respect to \(x\), we can take it out of the differentiation:

  \[ p = (y^2 + b) \frac{\delta}{\delta x}(x^2 + a) \]

  Differentiating \(x^2\) with respect to \(x\) gives \(2x\), and differentiating the constant \(a\) gives \(0\).

  \[ p = (y^2 + b) (2x + 0) \] \[ p = 2x(y^2 + b) \]
• Partial derivative of z with respect to y (\(\frac{\delta z}{\delta y}\) or \(q\)):

  Similarly, when differentiating with respect to \(y\), we treat \(x\) and the constant \(a\) as constants.

  \[ q = \frac{\delta z}{\delta y} = \frac{\delta}{\delta y} [(x^2 + a)(y^2 + b)] \]

  Since \((x^2 + a)\) is a constant with respect to \(y\), we can take it out of the differentiation:

  \[ q = (x^2 + a) \frac{\delta}{\delta y}(y^2 + b) \]

  Differentiating \(y^2\) with respect to \(y\) gives \(2y\), and differentiating the constant \(b\) gives \(0\).

  \[ q = (x^2 + a) (2y + 0) \] \[ q = 2y(x^2 + a) \]

Constants Elimination: Forming the Partial Differential Equation

Now we have the following two equations from our partial derivatives:

1. \(p = 2x(y^2 + b)\)
2. \(q = 2y(x^2 + a)\)

We need to use these to eliminate the arbitrary constants 'a' and 'b' from the original relation \(z = (x^2 + a)(y^2 + b)\).

• From equation (1), we can express \((y^2 + b)\): \[ y^2 + b = \frac{p}{2x} \]
• From equation (2), we can express \((x^2 + a)\): \[ x^2 + a = \frac{q}{2y} \]

Now, substitute these expressions for \((x^2 + a)\) and \((y^2 + b)\) back into the original relation \(z = (x^2 + a)(y^2 + b)\):

To simplify and rearrange this into the standard form of a partial differential equation, multiply both sides by \(4xy\):

Finally, substitute back the original notation for \(p\) and \(q\): \(p = \frac{\delta z}{\delta x}\) and \(q = \frac{\delta z}{\delta y}\).

Equation Summary: The Resulting Partial Differential Equation

The partial differential equation formed from the given relation \(z = (x^2 + a)(y^2 + b)\) is \(\frac{\delta z}{\delta x} \frac{\delta z}{\delta y} = 4xyz\). This process demonstrates how to derive a PDE by eliminating arbitrary constants through partial differentiation.