Find the general solution of x (∂ u)/(∂ x) + (z+u) (∂ u)/(∂…
Find the general solution of \[ x\,\frac{\partial u}{\partial x} + (z+u)\,\frac{\partial u}{\partial y} + (y+u)\,\frac{\partial u}{\partial z} = y + z, \] where $u = u(x,y,z)$.
Explanation of Method
A first-order linear PDE of Lagrange type in four variables $(x,y,z,u)$: $P=x$, $Q=z+u$, $R=y+u$, $S=y+z$. The first integral follows from summing the 2nd–4th ratios (since $Q+R+S=2(y+z+u)$); the second from their pairwise difference.
Step 1 · Characteristic equations
\[ \frac{dx}{x} = \frac{dy}{z+u} = \frac{dz}{y+u} = \frac{du}{y+z}. \tag{1} \]
Step 2 · First integral
Summing the 2nd, 3rd, 4th numerators and denominators: \[ \frac{dy+dz+du}{(z+u)+(y+u)+(y+z)} = \frac{d(y+z+u)}{2(y+z+u)}. \] Equating to the first ratio $dx/x$ and integrating: \[ \frac{y+z+u}{x^2} = c_1. \tag{I} \]
Step 3 · Second integral
Subtracting the 3rd ratio from the 2nd: \[ \frac{dy-dz}{(z+u)-(y+u)} = \frac{d(y-z)}{z-y} = -\frac{d(y-z)}{y-z}. \] Equating to $dx/x$ and integrating: \[ x(y-z) = c_2. \tag{II} \]
Step 4 · General solution
Since (I) involves $u$ while (II) does not, the two integrals are independent. The general solution $c_1=\Psi(c_2)$ is: \[ \frac{y+z+u}{x^2} = \Psi\!\left(x(y-z)\right), \] where $\Psi$ is an arbitrary differentiable function.
Verification
The condition for $F(x,y,z,u)$ to be a first integral of (1) is
$PF_x+QF_y+RF_z-SF_u=0$.
Check $c_1=(y+z+u)/x^2$: $(c_1)_x=-2(y+z+u)/x^3$, $(c_1)_y=(c_1)_z=(c_1)_u=1/x^2$:
\[ P(c_1)_x+Q(c_1)_y+R(c_1)_z-S(c_1)_u
= \frac{-2(y+z+u)+(z+u)+(y+u)-(y+z)}{x^2} = 0. \checkmark \]
Check $c_2=x(y-z)$: $(c_2)_x=y-z$, $(c_2)_y=x$, $(c_2)_z=-x$, $(c_2)_u=0$:
\[ P(c_2)_x+Q(c_2)_y+R(c_2)_z = x\!\left[(y-z)+(z+u)-(y+u)\right] = 0. \checkmark \]
Since both first integrals satisfy this condition, every surface
$\Psi(c_1,c_2)=0$ solves the PDE. $\checkmark$
Result
The general solution is: