General integral for partial differential equation I want to find a general integral for $xz_x + yz_y = z^2 + 1$, where $z = z(x,y)$ is implicitly a function of two variables. I'm not sure how to go about this.
 A: $$\dfrac {dx}{x}=\dfrac {dy}{y}=\dfrac {dz}{z^2+1}$$
First DE gives us:
$$d \ln x = d\ln y \implies \dfrac x y=C_1$$
Second DE gives us:
$$ d\ln y=d \arctan z$$
$$\ln y = \arctan z +C_2$$
The solution is therefore:
$$C_2= f(C_1)$$
$$\ln y - \arctan z = f\left (\dfrac x y \right)$$
$$ z (x,y)=\tan \left(\ln y+ f\left (\dfrac x y \right) \right)$$
A: Elaborating on my comment, the complete Lagrange-Charpit equations read
$$  \frac{\mathrm{d}x}{x} = \frac{\mathrm{d}y}{y} = \frac{\mathrm{d}z}{1+z^2} = \frac{\mathrm{d}p}{(2z-1)p} = \frac{\mathrm{d}q}{(2z-1)q}, $$
with $p=z_x$ and $q=z_y$. As in @Aryadeva's answer, from the first equality we get the characteristic $x/y=c_1$. If you then go with the equation formed with the 2nd and 3rd fraction, you obtain their solution. Alternatively, the 1st and 3rd would give you $z = \tan\left[ \log x + f(x/y) \right]$.
In order to obtain a complete integral you would need to use the 3rd and either 4th or 5th equation together with the fact that
$$\mathrm{d} z = p \mathrm{d}x + q \mathrm{d}y. \quad(1)$$
Let me use for instance 4th and 5th fractions to get
\begin{equation}
\frac{\mathrm{d}p}{(2z-1)p} = \frac{\mathrm{d}q}{(2z-1)q} \implies \frac{\mathrm{d}p}{p} = \frac{\mathrm{d}q}{q} \implies q = a_1 p,
\end{equation}
where $a_1$ is a constant of integration. Let us now use the 3rd and 4th fractions to arrive at
$$ \frac{2z-1}{z^2+1} \mathrm{d}z = \mathrm{d}z \implies p = a_2 (1+z^2)\mathrm{e}^{-\operatorname{atan}{z}},$$
with $a_2$ another constant of integration. Combine this into $(1)$ to finally get at
$$ \mathrm{d}z = p (\mathrm{d}x + a_1 \mathrm{d}y) = a_2 (1+z^2)\mathrm{e}^{-\operatorname{atan}{z}} (\mathrm{d}x + a_1 \mathrm{d}y), $$
or alternatively
$$  \frac{\mathrm{e}^{\operatorname{atan}{z}}}{z^2+1} \mathrm{d}z = \mathrm{d}\left( \mathrm{e}^{\operatorname{atan}{z}}  \right) = \mathrm{d}(A x + B y),$$ where I have renamed the constants $a_2=A$ and $a_1a_2=B$. The result in my comment follows upon integration and putting the new constant of integration as a function of $c_1 = x/y$, i.e.,
$$z = \tan\left[ \log(Ax+By) + f(x/y) \right]$$
