Find the integral surface of y^2 p + xy q = x, p=(∂ z)/(∂…

Find the integral surface of \[ y^2\,p + xy\,q = x, \qquad p=\frac{\partial z}{\partial x},\quad q=\frac{\partial z}{\partial y}, \] that passes through the curve $x=0$, $z=y^2$.

Step 1 · Characteristic equations and first integral

\[ \frac{dx}{y^2} = \frac{dy}{xy} = \frac{dz}{x}. \tag{1} \] From the first two ratios: $x\,dx=y\,dy \Rightarrow x^2-y^2=c_1$. \tag{I}

Step 2 · Second integral

From $x\,dx=y^2\,dz$ with $y^2=x^2-c_1$: $x\,dx=(x^2-c_1)\,dz$, so $dz=x\,dx/(x^2-c_1)$. Integrating: $z=\tfrac12\ln|x^2-c_1|+c_2=\ln|y|+c_2$, so \[ c_2 = z - \ln|y|. \tag{II} \]

Step 3 · General solution

\[ z = \ln|y| + \Phi(x^2-y^2). \tag{$*$} \]

Step 4 · Apply the initial condition

Set $x=0$, $z=y^2$: $y^2=\ln|y|+\Phi(-y^2)$, so with $t=-y^2$, $\Phi(t)=-t-\tfrac12\ln(-t)$ for $t<0$. Substituting $t=x^2-y^2$: \[ \Phi(x^2-y^2) = y^2-x^2 - \tfrac12\ln(y^2-x^2). \] Combining with ($*$) and using $\ln|y|=\tfrac12\ln y^2$: \[ z = y^2-x^2 + \frac{1}{2}\ln\frac{y^2}{y^2-x^2}. \]

Verification

At $x=0$: $z=y^2+\tfrac12\ln1=y^2$. $\checkmark$
Compute $p=-2x+\dfrac{x}{y^2-x^2}$, $q=2y+\dfrac1y-\dfrac{y}{y^2-x^2}$: \begin{align*} y^2p+xyq &= -2xy^2+\frac{xy^2}{y^2-x^2} + 2xy^2+x-\frac{xy^2}{y^2-x^2} = x. \checkmark \end{align*}