Find the integral surface of x(x^2+y^2) p + 2y^2(xp+yq-z) =…

Find the integral surface of \[ x(x^2+y^2)\,p + 2y^2(xp+yq-z) = 0, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}, \] that passes through the curve $z=h$, $x^2+y^2=a^2$.

Step 1 · Rewrite the PDE

Expanding: $x(x^2+y^2)p+2xy^2p+2y^3q-2y^2z=0$, i.e. \[ x(x^2+3y^2)\,p + 2y^3\,q = 2y^2z. \] Characteristic equations: \[ \frac{dx}{x(x^2+3y^2)} = \frac{dy}{2y^3} = \frac{dz}{2y^2z}. \tag{1} \]

Step 2 · First integral

From the last two ratios, $2y^2$ cancels: $dy/y=dz/z \Rightarrow z/y=c_1$. \tag{I}

Step 3 · Second integral

From the first two ratios, with $w=x^2$ as a function of $y$ along a characteristic: $dw/dy = w(w+3y^2)/y^3$, a Bernoulli equation. Setting $v=1/w=1/x^2$: \[ \frac{dv}{dy} = -\frac{1}{y^3} - \frac{3v}{y}, \] a linear ODE. With integrating factor $y^3$: $\dfrac{d(vy^3)}{dy}=-1$, so $vy^3=-y+c_2$, i.e. \[ c_2 = \frac{y^3}{x^2} + y. \tag{II} \]

Step 4 · General solution

\[ \frac{y^3}{x^2} + y = \Phi\!\left(\frac{z}{y}\right). \tag{$*$} \]

Step 5 · Apply the initial condition

Parametrise $z=h$, $x^2+y^2=a^2$ by $t=y$: $x^2=a^2-t^2$, $z=h$. Then $c_1=h/t$ and $c_2=a^2t/(a^2-t^2)$. Expressing $t=h/c_1$ and substituting gives $\Phi(s)=a^2hs/(a^2s^2-h^2)$. Substituting $s=z/y$ into ($*$) and clearing denominators: \[ (x^2+y^2)(a^2z^2-h^2y^2) = a^2hx^2z. \]

Verification

On $x^2+y^2=a^2$, $z=h$: both sides reduce to $a^2h^2(a^2-y^2)$. $\checkmark$ Since $c_1=z/y$ and $c_2=y^3/x^2+y$ both satisfy $P(c)_x+Q(c)_y+R(c)_z=0$ for the rewritten PDE, every surface $c_2=\Phi(c_1)$ solves it. $\checkmark$