Find the complete integral of yzp^2 = q

Find the complete integral of \[ yzp^2 = q. \]

Step 1 · Substitute $z=X(x)Y(y)$

With $p=X'Y$, $q=XY'$, the equation $yz p^2=q$ becomes \[ y\,XY\,(X'Y)^2 = XY' \implies yY^3(X')^2 = \frac{Y'}{Y}. \] For the left side to depend on $y$ alone (matching the right side), $(X')^2$ must be constant: set $X'=a$, so $X=ax$.

Step 2 · Solve the Bernoulli ODE for $Y$

With $X'=a$: \[ Y' = a^2y\,Y^3 \implies \frac{dY}{Y^3}=a^2y\,dy \implies -\frac{1}{2Y^2} = \frac{a^2y^2}{2}+C_0. \] Writing $b=-2C_0$: \[ Y = \frac{1}{\sqrt{b-a^2y^2}}. \]

Step 3 · Write the complete integral

\[ z = XY = \frac{ax}{\sqrt{b-a^2y^2}}. \]

Verification

\[ p = \frac{a}{\sqrt{b-a^2y^2}}, \qquad q = \frac{a^3xy}{(b-a^2y^2)^{3/2}}. \] \[ yzp^2 = y\cdot\frac{ax}{\sqrt{b-a^2y^2}}\cdot\frac{a^2}{b-a^2y^2} = \frac{a^3xy}{(b-a^2y^2)^{3/2}} = q. \checkmark \]