Find the general solution of x^2 z p + y^2 z q = x + y, p =…

Find the general solution of \[ x^2 z\,p + y^2 z\,q = x + y, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]

Step 1 · Characteristic equations

\[ \frac{dx}{x^2 z} = \frac{dy}{y^2 z} = \frac{dz}{x+y}. \tag{1} \]

Step 2 · First integral

From $\dfrac{dx}{x^2z}=\dfrac{dy}{y^2z}$, the factor $z$ cancels: $dx/x^2=dy/y^2$. Integrating: $-1/x=-1/y+\text{const}$, so \[ \frac{1}{y} - \frac{1}{x} = c_1. \tag{I} \]

Step 3 · Second integral

Apply multipliers $(1/x^2,1/y^2,0)$: the denominator becomes $\tfrac{1}{x^2}\cdot x^2z+\tfrac{1}{y^2}\cdot y^2z=2z$, and since $dx/x^2=-d(1/x)$, $dy/y^2=-d(1/y)$, the numerator is $-d(1/x+1/y)$. Thus \[ \frac{-d(1/x+1/y)}{2z} = \frac{dz}{x+y} \;\Longrightarrow\; -(x+y)\,d\!\left(\frac1x+\frac1y\right) = 2z\,dz. \] Let $u=1/x+1/y$; since $c_1=1/y-1/x$ is fixed, $x=2/(u-c_1)$, $y=2/(u+c_1)$, so $x+y=4u/(u^2-c_1^2)$. Substituting: \[ -\frac{2u\,du}{u^2-c_1^2} = z\,dz. \] Integrating: $-\ln(u^2-c_1^2)=\tfrac{z^2}{2}+\text{const}$. Since $u^2-c_1^2=\left(\tfrac1x+\tfrac1y\right)^2-\left(\tfrac1y-\tfrac1x\right)^2=\dfrac{4}{xy}$, \[ \frac{z^2}{2} - \ln(xy) = c_2, \tag{II} \] after absorbing $\ln 4$ into the constant.

Step 4 · General solution

The general solution is $c_2=\Phi(c_1)$: \[ \frac{z^2}{2} - \ln(xy) = \Phi\!\left(\frac{1}{y}-\frac{1}{x}\right), \] where $\Phi$ is an arbitrary differentiable function.

Verification

First integral: $c_{1x}=1/x^2$, $c_{1y}=-1/y^2$: $Pc_{1x}+Qc_{1y}=x^2z/x^2-y^2z/y^2=z-z=0. \checkmark$
Second integral: with $c_2=z^2/2-\ln(xy)$, $c_{2x}=-1/x$, $c_{2y}=-1/y$, $c_{2z}=z$: \[ Pc_{2x}+Qc_{2y}+Rc_{2z} = -xz-yz+(x+y)z = 0. \checkmark \] PDE check: differentiating $z^2/2-\ln(xy)=\Phi(1/y-1/x)$ with respect to $x$ and $y$ gives $zp-1/x=\Phi'/x^2$ and $zq-1/y=-\Phi'/y^2$; multiplying by $x^2z$ and $y^2z$ respectively and adding eliminates $\Phi'$: \[ x^2z^2p - xz + y^2z^2q - yz = 0 \;\Longrightarrow\; x^2zp+y^2zq = x+y. \checkmark \]