Question

find the integral surface of the equation : (x-y)y^2p + (y- x)x^2q=(x^2 + y^2)z,
through the curve xz =a^2, y = 0

Answer 1

$$Solution$$

The given partial differential equation is

$$(x-y)y^2P+(y-x)x^2q=(x^2+y^2)z$$

The Lagrange’s auxiliary equations for given PDE is

$$\frac{dx}{(x-y)\cdot y^2}=\frac{dy}{(y-x)\cdot x^2}=\frac{dz}{(x^2+y^2) \cdot z}\\ Taking\ dx\ and\ dy\\ \frac{dx}{(x-y)\cdot y^2}=\frac{dy}{(y-x) \cdot x^2}$$

$\frac{dy}{dx}=-\frac{x^2}{y^2}$

$dy \cdot y^2=dx \cdot (-x^2)$

$C_1=y^3+x^3------(i)$

$\frac{dx}{(x-y) \cdot y^2}=\frac{dz}{(x^2+y^2)\cdot z}$

$\frac{dz}{z}=\frac{dx(x^2+y^2)}{(x-y) \cdot y^2}$

$lnz=\frac{1}{y^2}\cdot F(x,y)+C_2--(ii)$

Here, y cannot be 0 in this equation that is why there is no intersection with curve $xz=a^2, y=0.$

Answer: No solution.