Find the complete and singular integrals of p^2x^2 = z(z-qy)
Find the complete and singular integrals of \[ p^2x^2 = z(z-qy). \]
Explanation of Method
Divide by $z^2$ and substitute $u=\ln z$ to remove the explicit $z$-dependence, reducing to a separable equation in $u_x,u_y$.
Step 1 · Substitute $u=\ln z$
Dividing by $z^2$: $p^2x^2/z^2=1-qy/z$. With $z=e^u$, $p/z=u_x$, $q/z=u_y$: \[ u_x^2x^2+u_yy = 1. \]
Step 2 · Separate variables
Write $u_x^2x^2=1-u_yy$ and set both sides equal to $a$: \[ u_x = \frac{\sqrt a}{x}, \qquad u_y = \frac{1-a}{y}. \]
Step 3 · Integrate
\[ u = \sqrt a\ln|x|+(1-a)\ln|y|+b \implies z = b\,x^{\sqrt a}y^{1-a}. \]
Step 4 · Find the singular integral
With $F=p^2x^2-z^2+qyz$: $F_p=2px^2=0 \Rightarrow p=0$; substituting $p=0$ into the original equation gives $0=z(z-qy)$, and the degenerate branch consistent with $F_q=-yz=0$ gives \[ z = 0. \]
Verification
With $z=bx^{\sqrt a}y^{1-a}$: $p=\sqrt a\,z/x$, $q=(1-a)z/y$, so \[ p^2x^2 = az^2, \qquad z(z-qy) = z^2(1-(1-a)) = az^2. \checkmark \]
Result
The complete and singular integrals are: