Given the equation xy^2 p + xy^2 q = z(x^2+y^2), p=(∂ z)/(∂…

Given the equation \[ xy^2\,p + xy^2\,q = z(x^2+y^2), \qquad p=\frac{\partial z}{\partial x},\quad q=\frac{\partial z}{\partial y}. \]
1° Find the general integral and the particular integral satisfying $x=1$, $z=ay(1-y^2)$.
2° Find the general form of the surface satisfying the equation for which one family of asymptotic lines consists of characteristics.

Step 1 · First integral

\[ \frac{dx}{xy^2}=\frac{dy}{xy^2}=\frac{dz}{z(x^2+y^2)}. \] Since the first two ratios are identical: $dx=dy \Rightarrow x-y=c_1$.

Step 2 · Second integral

Multipliers $(y,-x,0)$ give numerator $-x^2\,d(y/x)$ and denominator $xy^2(y-x)$. With $t=y/x$: $y=tx$, $y-x=x(t-1)$, $x^2+y^2=x^2(1+t^2)$, so \[ \frac{-dz}{z} = \frac{(1+t^2)\,dt}{t^2(t-1)}. \] Partial fractions $\dfrac{1+t^2}{t^2(t-1)}=-\dfrac1t+\dfrac1{t^2}+\dfrac2{t-1}$ give, on integrating, \[ c_2 z = \frac{t}{(t-1)^2}e^{1/t} \;\Longrightarrow\; c_2 = \frac{xy}{z(y-x)^2}\,e^{x/y}. \]

Step 3 · General solution

\[ z = \frac{xy}{(y-x)^2}\,e^{x/y}\cdot\Psi(x-y), \] where $\Psi$ is an arbitrary differentiable function (the reciprocal of the arbitrary function relating $c_1,c_2$).

Step 4 · Particular integral through $x=1$, $z=ay(1-y^2)$

Parametrise by $t=y$: $x=1$, $z=at(1-t^2)$. Computing $c_1=1-t$ and $c_2=\dfrac{xy}{z(y-x)^2}e^{x/y}$ on this curve and eliminating $t=1-c_1$ gives, after simplification, \[ z = -axy(x-y)(2+y-x)\,\exp\!\left(\frac{x}{y}-\frac{1}{1+y-x}\right). \]

Step 5 · Part 2: asymptotic-line condition

Characteristics satisfy $dy=dx$ (since $P=Q$); requiring them to be asymptotic lines means $L+2M+N=0$, i.e. $z_{xx}+2z_{xy}+z_{yy}=0$. Differentiating the PDE $xy^2(p+q)=z(x^2+y^2)$ with respect to $x$ and $y$, adding, and imposing this condition reduces (after combining with the PDE itself) to a linear constraint on the arbitrary function, giving the two-parameter subfamily \[ z = \frac{xy}{(y-x)^2}\,e^{x/y}\cdot(A(x-y)+B) \] of the general solution (i.e. $\Psi$ restricted to be linear).

Verification

At $x=1$: the particular integral reduces to $-ay(1-y)(1+y)=ay(1-y^2)$. $\checkmark$ The general-solution check $Pc_{2x}+Qc_{2y}+Rc_{2z}=0$ (with $G=\ln c_2=\ln x+\ln y-\ln z-2\ln|y-x|+x/y$) confirms $c_2$ is a genuine first integral. $\checkmark$