Find the complete and singular integrals of…
Find the complete and singular integrals of \[ 16p^2z^2+9q^2z^2+4z^2-4 = 0. \]
Explanation of Method
Substitute $w=z^2$ to remove the explicit $z$-factor, then use a travelling-wave reduction $w=f(ax+by)$ with a normalisation on $a,b$ forced by the coefficients $16,9$.
Step 1 · Substitute $w=z^2$
With $p=w_x/(2\sqrt w)$, $q=w_y/(2\sqrt w)$: \[ 16w_x^2+9w_y^2 = 16(1-w). \]
Step 2 · Travelling-wave reduction
With $w=f(ax+by)+c$, separating and integrating gives, after normalising $16a^2+9b^2=4$ (so $b=\tfrac23\sqrt{1-4a^2}$, $|a|\le\tfrac12$), \[ w = 1-(ax+by+c)^2 \implies z^2 = 1-(ax+by+c)^2. \]
Step 3 · Find the singular integral
$F_p=32pz^2=0$, $F_q=18qz^2=0$ give $p=q=0$ ($z\neq0$); substituting into the original equation gives $4z^2=4$: \[ z = \pm1. \]
Verification
From $z^2=1-(ax+by+c)^2$: $p=-a(\cdots)/z$, $q=-b(\cdots)/z$, so \[ 16p^2z^2+9q^2z^2 = (16a^2+9b^2)(ax+by+c)^2 = 4(ax+by+c)^2, \] and $16p^2z^2+9q^2z^2+4z^2 = 4(ax+by+c)^2+4[1-(ax+by+c)^2]=4$. \checkmark
Result
The complete and singular integrals are: