Find the complete and singular integrals of…

Find the complete and singular integrals of \[ pq+2x(y+1)p+y(y+2)q-2(y+1)z = 0. \]

Step 1 · Introduce $w=z+xy(y+2)$

Since $(p+y(y+2))(q+2x(y+1))=pq+2x(y+1)p+y(y+2)q+2xy(y+1)(y+2)$, the equation becomes, with $P=w_x=p+y(y+2)$, $Q=w_y=q+2x(y+1)$: \[ PQ = 2(y+1)\,w. \]

Step 2 · Substitute $u=2\sqrt w$

With $w=u^2/4$, $P=uu_x/2$, $Q=uu_y/2$: \[ u_xu_y = 2(y+1). \]

Step 3 · Separate and integrate

Set $u_x=a$, so $u=ax+g(y)$; then $ag'(y)=2(y+1)\Rightarrow g(y)=(y+1)^2/a+c$: \[ u = ax+\frac{(y+1)^2}{a}+c. \]

Step 4 · Write the complete integral

Squaring $u=2\sqrt w$ and using $w=z+xy(y+2)$: \[ z = \frac14\!\left(ax+\frac{(y+1)^2}{a}+c\right)^{\!2}-xy(y+2). \]

Step 5 · Find the singular integral

$F_p=q+2x(y+1)=0$, $F_q=p+y(y+2)=0$ give $q=-2x(y+1)$, $p=-y(y+2)$; substituting into the original equation and simplifying: \[ z = -xy(y+2). \]

Verification

Let $\varphi=ax+(y+1)^2/a+c$. Then $p+y(y+2)=a\varphi/2$ and $q+2x(y+1)=(y+1)\varphi/a$, so \[ (p+y(y+2))(q+2x(y+1)) = \frac{(y+1)\varphi^2}{2} = 2(y+1)\cdot\frac{\varphi^2}{4}=2(y+1)w. \checkmark \] Singular: substituting $p=-y(y+2)$, $q=-2x(y+1)$ into the original equation gives $-2xy(y+1)(y+2)-2(y+1)z=0 \Rightarrow z=-xy(y+2)$. \checkmark