Find the general solution of (s-x) (∂ u)/(∂ x) +(s-y) (∂…

Find the general solution of \[ (s-x)\,\frac{\partial u}{\partial x} +(s-y)\,\frac{\partial u}{\partial y} +(s-z)\,\frac{\partial u}{\partial z} = s - u, \qquad s = x+y+z+u. \]

Step 1 · Characteristic equations

With $P=y+z+u$, $Q=x+z+u$, $R=x+y+u$, $S=x+y+z$: \[ \frac{dx}{y+z+u} = \frac{dy}{x+z+u} = \frac{dz}{x+y+u} = \frac{du}{x+y+z}. \tag{1} \]

Step 2 · Each variable obeys the same ODE in $s$

Summing all four ratios: numerator $dx+dy+dz+du=ds$, denominator $3(x+y+z+u)=3s$, so every ratio in (1) equals $ds/(3s)$. In particular, \[ \frac{dx}{ds} = \frac{s-x}{3s} = \frac{1}{3}-\frac{x}{3s}, \] a first-order linear ODE. With integrating factor $s^{1/3}$: \[ \frac{d}{ds}\!\left(xs^{1/3}\right) = \frac{s^{1/3}}{3} \;\Longrightarrow\; xs^{1/3} = \frac{s^{4/3}}{4}+A \;\Longrightarrow\; \left(x-\frac{s}{4}\right)s^{1/3}=A. \] By complete symmetry, the identical ODE holds for $y$, $z$, $u$, so $(y-s/4)s^{1/3}$, $(z-s/4)s^{1/3}$, $(u-s/4)s^{1/3}$ are each constant along a characteristic — but their sum is identically $0$, so only three of the four are independent.

Step 3 · Three independent first integrals from differences

Since the $s/4$ terms cancel in any difference, pairwise differences give \[ (x-y)\,s^{1/3} = c_1, \qquad (y-z)\,s^{1/3} = c_2, \qquad (z-u)\,s^{1/3} = c_3. \tag{I--III} \] These three combinations are functionally independent (no one is a combination of the other two).

Step 4 · General solution

The general solution $c_3=\Psi(c_1,c_2)$ is: \[ (z-u)(x+y+z+u)^{1/3} = \Psi\!\left((x-y)(x+y+z+u)^{1/3},\,(y-z)(x+y+z+u)^{1/3}\right), \] where $\Psi$ is an arbitrary differentiable function of two arguments.

Verification

The condition for $F(x,y,z,u)$ to be a first integral of (1) is $PF_x+QF_y+RF_z+SF_u=0$. Let $\sigma=s^{1/3}$, so $\sigma_x=\sigma_y=\sigma_z=\sigma_u=\tfrac13 s^{-2/3}$. For $c_1=(x-y)\sigma$: $(c_1)_x=\sigma+c_1/(3s)$, $(c_1)_y=-\sigma+c_1/(3s)$, $(c_1)_z=(c_1)_u=c_1/(3s)$, so \begin{align*} Pc_{1x}+Qc_{1y}+Rc_{1z}+Sc_{1u} &= \sigma\!\left[(y+z+u)-(x+z+u)\right] + \frac{c_1}{3s}\!\left[3s\right]\\ &= \sigma(y-x)+c_1 = -c_1+c_1 = 0. \checkmark \end{align*} By complete symmetry the same calculation confirms $c_2=(y-z)\sigma$ and $c_3=(z-u)\sigma$ are first integrals. $\checkmark$