Find the general solution of (u-x) (∂ u)/(∂ x) +(u-y) (∂…
Find the general solution of \[ (u-x)\,\frac{\partial u}{\partial x} +(u-y)\,\frac{\partial u}{\partial y} - z\,\frac{\partial u}{\partial z} = x+y, \] where $u = u(x,y,z)$.
Explanation of Method
A Lagrange PDE with $P=u-x$, $Q=u-y$, $R=-z$, $S=x+y$. The first integral follows from the difference of the first two ratios paired with the third; the second from the combination $(1,1,-1)$ on $x,y,u$ paired with the third ratio; the third from an auxiliary linear ODE obtained by eliminating $u$ using the second integral.
Step 1 · Characteristic equations
\[ \frac{dx}{u-x} = \frac{dy}{u-y} = \frac{dz}{-z} = \frac{du}{x+y}. \tag{1} \]
Step 2 · First integral
Subtracting the first two ratios: $\dfrac{dx-dy}{(u-x)-(u-y)} = \dfrac{d(x-y)}{y-x} = -\dfrac{d(x-y)}{x-y}$. Equating to the third ratio $dz/(-z)$: \[ \frac{d(x-y)}{x-y} = \frac{dz}{z} \;\Longrightarrow\; \frac{x-y}{z} = c_1. \tag{I} \]
Step 3 · Second integral
Combine the first, second, and fourth ratios with multipliers $(1,1,-1)$: \[ \frac{dx+dy-du}{(u-x)+(u-y)-(x+y)} = \frac{d(x+y-u)}{-2(x+y-u)}. \] Equating to the third ratio and simplifying: \[ \frac{d(x+y-u)}{2(x+y-u)} = \frac{dz}{z} \;\Longrightarrow\; \frac{x+y-u}{z^2} = c_2. \tag{II} \]
Step 4 · Third integral
Let $w=x+y+u$. Summing the first, second, fourth ratios with multipliers $(1,1,1)$ and equating to the third ratio: $dw/(2u)=dz/(-z)$, i.e. $dw=-2u\,dz/z$. From (II), $x+y-u=c_2z^2$, so $u=(w-c_2z^2)/2$. Substituting gives the linear ODE \[ \frac{dw}{dz} + \frac{w}{z} = c_2 z. \] With integrating factor $z$: $\dfrac{d(wz)}{dz}=c_2z^2 \Rightarrow wz = \dfrac{c_2z^3}{3}+c_3$. Substituting back $w=x+y+u$ and $c_2=(x+y-u)/z^2$ and simplifying: \[ c_3 = z(x+y+u) - \frac{(x+y-u)z}{3} = \frac{z\bigl[2(x+y)+4u\bigr]}{3}, \] so, absorbing the constant factor $2/3$, \[ z(x+y+2u) = c_3. \tag{III} \]
Step 5 · General solution
The general solution $c_1=\Psi(c_2,c_3)$ is: \[ \frac{x-y}{z} = \Psi\!\left(\frac{x+y-u}{z^2},\;z(x+y+2u)\right), \] where $\Psi$ is an arbitrary differentiable function of two arguments.
Verification
The condition for $F(x,y,z,u)$ to be a first integral is $PF_x+QF_y+RF_z+SF_u=0$.
$c_1=(x-y)/z$: $(c_1)_x=1/z$, $(c_1)_y=-1/z$, $(c_1)_z=-(x-y)/z^2$:
\[ \frac{u-x}{z}-\frac{u-y}{z}+\frac{(x-y)z}{z^2} = \frac{y-x}{z}+\frac{x-y}{z}=0. \checkmark \]
$c_2=(x+y-u)/z^2$: $(c_2)_x=(c_2)_y=1/z^2$, $(c_2)_z=-2(x+y-u)/z^3$, $(c_2)_u=-1/z^2$:
\[ Pc_{2x}+Qc_{2y}+Rc_{2z}+Sc_{2u}
= \frac{(u-x)+(u-y)}{z^2} + \frac{2(x+y-u)}{z^2} - \frac{x+y}{z^2}
= \frac{2u-x-y+2x+2y-2u-x-y}{z^2} = 0. \checkmark \]
$c_3=z(x+y+2u)$: $(c_3)_x=(c_3)_y=z$, $(c_3)_z=x+y+2u$, $(c_3)_u=2z$:
\[ (u-x)z+(u-y)z-z(x+y+2u)+2z(x+y) = z\!\left[2u-x-y-x-y-2u+2x+2y\right]=0. \checkmark \]
Result
The general solution is: