Partial Differential Equation Analysis

The given partial differential equation is \(z=x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}+\frac{\partial z}{\partial x} \frac{\partial z}{\partial y}\).

Let \(p = \frac{\partial z}{\partial x}\) and \(q = \frac{\partial z}{\partial y}\). The equation can be written as:

\(z = xp + yq + pq\)

This equation is in the form \(z = px + qy + f(p, q)\), which is known as a Clairaut's equation in partial differential equations. For a Clairaut's equation, the complete integral is obtained by simply replacing \(p\) with an arbitrary constant \(a\) and \(q\) with an arbitrary constant \(b\).

Complete Integral Derivation

For the given PDE, \(f(p, q) = pq\).

Replacing \(p\) with \(a\) and \(q\) with \(b\), the complete integral is:

\(z = ax + by + ab\)

where \(a\) and \(b\) are arbitrary constants.

Let's examine the given options:

• Option 1: The complete integral is \(z = xa + yb + ab\), a, b arbitrary constants. This matches our derived complete integral.
• Option 2: The complete integral is \(z = xa + yb + \sqrt{a^2+b^2}\), a, b arbitrary constants. This does not match the form \(z = ax + by + ab\), so it is incorrect.

Based on this, Option 1 is a true statement.

Particular Solution Verification

We need to find or verify the particular solution passing through the condition \(x = 0\) and \(z = y^2\).

Let's check the proposed particular solutions from the options.

• Option 3 proposes the particular solution \(z = \left(\frac{x}{4}-y\right)^2\).

  First, let's check if this solution satisfies the initial condition \(z = y^2\) when \(x=0\).

  Substitute \(x=0\) into \(z = \left(\frac{x}{4}-y\right)^2\):

  \(z = \left(\frac{0}{4}-y\right)^2 = (-y)^2 = y^2\)

  This matches the condition \(z=y^2\) at \(x=0\).

  Next, let's check if this solution satisfies the original partial differential equation \(z = xp + yq + pq\).

  If \(z = \left(\frac{x}{4}-y\right)^2\), then:

  \(p = \frac{\partial z}{\partial x} = 2\left(\frac{x}{4}-y\right) \cdot \frac{1}{4} = \frac{1}{2}\left(\frac{x}{4}-y\right)\)

  \(q = \frac{\partial z}{\partial y} = 2\left(\frac{x}{4}-y\right) \cdot (-1) = -2\left(\frac{x}{4}-y\right)\)

  Substitute these into the PDE \(z = xp + yq + pq\):

  RHS \( = x \left[\frac{1}{2}\left(\frac{x}{4}-y\right)\right] + y \left[-2\left(\frac{x}{4}-y\right)\right] + \left[\frac{1}{2}\left(\frac{x}{4}-y\right)\right]\left[-2\left(\frac{x}{4}-y\right)\right]\)

  RHS \( = \frac{x}{2}\left(\frac{x}{4}-y\right) - 2y\left(\frac{x}{4}-y\right) - \left(\frac{x}{4}-y\right)^2\)

  Factor out \(\left(\frac{x}{4}-y\right)\):

  RHS \( = \left(\frac{x}{4}-y\right) \left[\frac{x}{2} - 2y - \left(\frac{x}{4}-y\right)\right]\)

  RHS \( = \left(\frac{x}{4}-y\right) \left[\frac{x}{2} - 2y - \frac{x}{4} + y\right]\)

  RHS \( = \left(\frac{x}{4}-y\right) \left[\left(\frac{x}{2} - \frac{x}{4}\right) + (-2y + y)\right]\)

  RHS \( = \left(\frac{x}{4}-y\right) \left[\frac{x}{4} - y\right]\)

  RHS \( = \left(\frac{x}{4}-y\right)^2\)

  This is equal to \(z\). So, the solution \(z = \left(\frac{x}{4}-y\right)^2\) satisfies the PDE and the initial condition. Thus, Option 3 is a true statement.

• Option 4 proposes the particular solution \(z = \left(\frac{x}{4}+y\right)^2\).

  Let's check if this solution satisfies the PDE \(z = xp + yq + pq\).

  If \(z = \left(\frac{x}{4}+y\right)^2\), then:

  \(p = \frac{\partial z}{\partial x} = 2\left(\frac{x}{4}+y\right) \cdot \frac{1}{4} = \frac{1}{2}\left(\frac{x}{4}+y\right)\)

  \(q = \frac{\partial z}{\partial y} = 2\left(\frac{x}{4}+y\right) \cdot (1) = 2\left(\frac{x}{4}+y\right)\)

  Substitute these into the PDE \(z = xp + yq + pq\):

  RHS \( = x \left[\frac{1}{2}\left(\frac{x}{4}+y\right)\right] + y \left[2\left(\frac{x}{4}+y\right)\right] + \left[\frac{1}{2}\left(\frac{x}{4}+y\right)\right]\left[2\left(\frac{x}{4}+y\right)\right]\)

  RHS \( = \frac{x}{2}\left(\frac{x}{4}+y\right) + 2y\left(\frac{x}{4}+y\right) + \left(\frac{x}{4}+y\right)^2\)

  RHS \( = \left(\frac{x}{4}+y\right) \left[\frac{x}{2} + 2y\right] + \left(\frac{x}{4}+y\right)^2\)

  RHS \( = \left(\frac{x}{4}+y\right) \left[\frac{x}{2} + 2y + \frac{x}{4} + y\right]\)

  RHS \( = \left(\frac{x}{4}+y\right) \left[\frac{3x}{4} + 3y\right]\)

  RHS \( = 3\left(\frac{x}{4}+y\right) \left(\frac{x}{4} + y\right) = 3\left(\frac{x}{4}+y\right)^2\)

  This is not equal to \(z = \left(\frac{x}{4}+y\right)^2\). So, Option 4 is incorrect.

Therefore, the true statements are Option 1 and Option 3.