Find the general solution of (x^2 + y^2)p + 2xy q + z^2 = 0…
Find the general solution of \[ \left(x^2 + y^2\right)p + 2xy\,q + z^2 = 0, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]
Explanation of Method
Rewritten as $(x^2+y^2)p+2xyq=-z^2$, a Lagrange PDE with $P=x^2+y^2$, $Q=2xy$, $R=-z^2$. The first integral follows from the first two ratios; the second uses the additive multipliers $(1,1,0)$, since $P+Q=(x+y)^2$ exactly matches the numerator $d(x+y)$, producing an exact differential in $1/z$ and $1/(x+y)$.
Step 1 · Characteristic equations
\[ \frac{dx}{x^2+y^2} = \frac{dy}{2xy} = \frac{dz}{-z^2}. \tag{1} \]
Step 2 · First integral
From $\dfrac{dx}{x^2+y^2}=\dfrac{dy}{2xy}$, cross-multiplying gives $2xy\,dx-(x^2+y^2)\,dy=0$, i.e. $2xy\,dx-x^2\,dy=y^2\,dy$. Dividing by $y^2$: \[ \frac{2xy\,dx-x^2\,dy}{y^2} = dy \;\Longrightarrow\; d\!\left(\frac{x^2}{y}\right)=dy. \] Integrating: $\dfrac{x^2}{y}=y+c_1$, i.e. \[ \frac{x^2-y^2}{y} = c_1. \tag{I} \]
Step 3 · Second integral
Apply multipliers $(1,1,0)$ to the three ratios: since $1\cdot(x^2+y^2)+1\cdot 2xy=(x+y)^2$, \[ \frac{dx+dy}{(x+y)^2} = \frac{d(x+y)}{(x+y)^2} = \frac{dz}{-z^2}. \] Cross-multiplying: $(x+y)^2\,dz + z^2\,d(x+y)=0$. Dividing by $z^2(x+y)^2$: \[ \frac{dz}{z^2} + \frac{d(x+y)}{(x+y)^2} = 0 \;\Longrightarrow\; d\!\left(-\frac{1}{z}\right) + d\!\left(-\frac{1}{x+y}\right) = 0 \;\Longrightarrow\; d\!\left(\frac{1}{z}+\frac{1}{x+y}\right)=0. \] Hence: \[ \frac{1}{z} + \frac{1}{x+y} = c_2. \tag{II} \]
Step 4 · General solution
The general solution is $c_2=\Phi(c_1)$: \[ \frac{1}{z} + \frac{1}{x+y} = \Phi\!\left(\frac{x^2-y^2}{y}\right), \] where $\Phi$ is an arbitrary differentiable function.
Verification
First integral: with $c_1=(x^2-y^2)/y$, $c_{1x}=2x/y$, $c_{1y}=-(x^2+y^2)/y^2$: \[ P c_{1x}+Qc_{1y} = (x^2+y^2)\frac{2x}{y} - 2xy\cdot\frac{x^2+y^2}{y^2} = \frac{2x(x^2+y^2)}{y}-\frac{2x(x^2+y^2)}{y} = 0. \checkmark \] Second integral: with $F=1/z+1/(x+y)$, $F_x=F_y=-1/(x+y)^2$, $F_z=-1/z^2$: \[ PF_x+QF_y+RF_z = -\frac{x^2+y^2+2xy}{(x+y)^2}+1 = -1+1=0. \checkmark \] PDE check: differentiating $1/z+1/(x+y)=\Phi(c_1)$ with respect to $x$ and $y$ and eliminating $\Phi'$ between the two relations (multiply and combine with $x^2+y^2$ and $2xy$ respectively) yields \[ (x^2+y^2)p+2xyq=-z^2. \checkmark \]
Result
The general solution is: