Solving the Differential Equation (xy p x-y q 1), (z 1), (y 2^{-1/2})

The given problem involves solving the differential equation with the Cauchy initial value problem xy p x-y q 1, z 1, and y 2-1/2. This problem can be addressed by employing the method based on the general integral and Lagrange's auxiliary equation.

Introduction to Lagrange's Auxiliary Equation

The method for solving the given equation begins with considering its general integral and Lagrange's auxiliary equation:

dx/xy dy/x-y dZ/1

This can be rewritten as a system of differential equations:

dx/ds xy dy/ds x-y dZ/ds 1

Integration and Integral Curves

From the first two equations, we can derive:

x-ydx - xydy 0

And similarly:

xdx - ydy - xdyydx 0

By integrating these, we obtain the integral:

x^2 - y^2 - 2xy C1

When we limit our attention to the region where y 0, this integral simplifies to:

y -x - 2x^2 - C11/2

Solving for z

Substituting the simplified integral into the original equations, we find:

dx/dZ 2x^2 - C11/2

Separating and integrating gives:

2-1/2ln[2x^2 - C11/2 21/2x] Z C2

Which can be rearranged as:

ln(xy21/2x - 21/2Z) C2

From this, we can generalize and obtain:

F(x^2 - y^2 - 2xy) ln(xy21/2x - 21/2Z) 0

Or equivalently:

ln(xy21/2x - 21/2Z) G(x^2 - y^2 - 2xy)

Applying the Cauchy Condition

By applying the Cauchy condition (z 1 and y 2-1/2) and substituting the values, we obtain:

ln(1 - 2y - 2y2) G(1 - y2 - 2y)

Introducing the variable V by setting 1 - y2 - 2y V, we find:

y -1 - V1/2

and

GV ln(2 - V1/2) - 1

Thus, the solution G(x^2 - y^2 - 2xy) ln(2 - x^2y^2 - 2xy1/2) - 1 is derived.

Final Solution

Substituting back, we get the solution for the given problem as:

21/2Zxy 1 - ln(xy21/2x) ln(2 - x^2y^2 - 2xy1/221/2) - 1

This provides a comprehensive approach to solving the given differential equation with the specified initial conditions.