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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.

  • You could try changing to polar coordinates in the $xy$-plane (since the left-hand side of the PDE is proportional to the directional derivative of $z$ in the radial direction). – Hans Lundmark Jan 01 '21 at 21:26
  • Using the Lagrange-Charpit method, a solution to your PDE is $z = \tan\left[\log x + f(x/y) \right]$, where $f$ is an arbitrary function. – Dmoreno Jan 01 '21 at 22:25
  • As a matter of fact, for $A, B \in \mathbb{R}$, $z = \tan \left[ \log(Ax +By) + f(x/y) \right]$ is a solution. – Dmoreno Jan 01 '21 at 22:41
  • @Dmoreno going over the Lagrange-Charpit method, I'm having trouble seeing how you arrived at that solution. Could you please elaborate? –  Jan 02 '21 at 02:06

2 Answers2

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$$\dfrac {dx}{x}=\dfrac {dy}{y}=\dfrac {dz}{z^2+1}$$ 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: $$C_2= f(C_1)$$ $$\ln y - \arctan z = f\left (\dfrac x y \right)$$ $$ z (x,y)=\tan \left(\ln y+ f\left (\dfrac x y \right) \right)$$

user577215664
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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.,

$$z = \tan\left[ \log(Ax+By) + f(x/y) \right]$$

Dmoreno
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