Find the general solution of x (∂ u)/(∂ x) + y (∂ u)/(∂ y) +…
Find the general solution of \[ x\,\frac{\partial u}{\partial x} + y\,\frac{\partial u}{\partial y} + (z+u)\,\frac{\partial u}{\partial z} = xy, \] where $u = u(x,y,z)$.
Explanation of Method
A Lagrange PDE with $P=x$, $Q=y$, $R=z+u$, $S=xy$. Introducing the parameter $\tau=\ln x$ turns the characteristic system into ordinary differential equations in $\tau$, which are solved successively for $y$, $u$, and finally $z$ (a first-order linear ODE, solved with an integrating factor).
Step 1 · Characteristic equations
\[ \frac{dx}{x} = \frac{dy}{y} = \frac{dz}{z+u} = \frac{du}{xy}. \tag{1} \] With $\tau=\ln x$ (so $x=e^\tau$), the system reads $\dot x=x$, $\dot y=y$, $\dot z=z+u$, $\dot u=xy$.
Step 2 · First integral
From $\dfrac{dx}{x}=\dfrac{dy}{y}$: $\dfrac{y}{x}=c_1$. \tag{I}
Step 3 · Second integral
From $\dfrac{dx}{x}=\dfrac{du}{xy}$: $du=y\,dx$. Using $y=c_1x$: $du=c_1x\,dx \Rightarrow u=\tfrac{c_1x^2}{2}+c_2$. Substituting $c_1=y/x$: \[ u - \frac{xy}{2} = c_2. \tag{II} \]
Step 4 · Third integral
With $u=c_2+c_1e^{2\tau}/2$, the ODE for $z$ becomes \[ \frac{dz}{d\tau} = z + c_2 + \frac{c_1e^{2\tau}}{2}. \] Multiplying by the integrating factor $e^{-\tau}$ and integrating: \[ ze^{-\tau} = -c_2e^{-\tau} + \frac{c_1e^{\tau}}{2} + c_3 \;\Longrightarrow\; z = -c_2 + \frac{c_1x^2}{2} + c_3x. \] Substituting $c_1=y/x$, $c_2=u-xy/2$: \[ z = -u + xy + c_3x \;\Longrightarrow\; c_3 = \frac{z+u-xy}{x}. \tag{III} \]
Step 5 · General solution
The general solution $c_3=\Psi(c_1,c_2)$ is: \[ \frac{z+u-xy}{x} = \Psi\!\left(\frac{y}{x},\; u-\frac{xy}{2}\right), \] where $\Psi$ is an arbitrary differentiable function.
Verification
The condition for $F(x,y,z,u)$ to be a first integral is
$PF_x+QF_y+RF_z+SF_u=0$.
$c_1=y/x$: $x(-y/x^2)+y(1/x)=0. \checkmark$
$c_2=u-xy/2$: $x(-y/2)+y(-x/2)+xy\cdot1 = -xy/2-xy/2+xy=0. \checkmark$
$c_3=(z+u-xy)/x$: with $(c_3)_x=-(z+u)/x^2$, $(c_3)_y=-1$, $(c_3)_z=1/x$,
$(c_3)_u=1/x$:
\[ x\!\left(-\frac{z+u}{x^2}\right) + y(-1) + (z+u)\cdot\frac{1}{x} + xy\cdot\frac{1}{x}
= -\frac{z+u}{x}-y+\frac{z+u}{x}+y = 0. \checkmark \]
Result
The general solution is: