Find the general solution of (y-bz) (∂ z)/(∂ x) - (x-az) (∂…
Find the general solution of \[ (y-bz)\,\frac{\partial z}{\partial x} - (x-az)\,\frac{\partial z}{\partial y} = bx - ay, \] where $a$ and $b$ are real constants.
Explanation of Method
A Lagrange PDE with $P=y-bz$, $Q=-(x-az)$, $R=bx-ay$. Remarkably, the multiplier combinations $(a,b,1)$ and $(x,y,z)$ both make the combined denominator vanish identically, so both first integrals are exact differentials of the numerator alone.
Step 1 · Characteristic equations
\[ \frac{dx}{y-bz} = \frac{dy}{-(x-az)} = \frac{dz}{bx-ay}. \tag{1} \]
Step 2 · First integral
Apply multipliers $(a,b,1)$: the combined denominator is $a(y-bz)+b(-(x-az))+(bx-ay) = ay-abz-bx+abz+bx-ay=0$, so $a\,dx+b\,dy+dz=0$ identically along every characteristic: \[ ax+by+z = c_1. \tag{I} \]
Step 3 · Second integral
Apply multipliers $(x,y,z)$: the combined denominator is $x(y-bz)-y(x-az)+z(bx-ay) = xy-bxz-xy+ayz+bxz-ayz=0$, so $x\,dx+y\,dy+z\,dz=0$: \[ x^2+y^2+z^2 = c_2. \tag{II} \]
Step 4 · General solution
The general solution $c_2=\Phi(c_1)$ is: \[ x^2+y^2+z^2 = \Phi(ax+by+z), \] where $\Phi$ is an arbitrary differentiable function.
Verification
Both multiplier identities directly confirm $P(c_1)_x+Q(c_1)_y+R(c_1)_z=0$ and $P(c_2)_x+Q(c_2)_y+R(c_2)_z=0$. For the PDE: differentiating $x^2+y^2+z^2=\Phi(ax+by+z)$ with respect to $x$ and $y$ gives $2x+2zp=\Phi'(a+p)$ and $2y+2zq=\Phi'(b+q)$; eliminating $\Phi'$ between these two relations and simplifying yields \[ (y-bz)p-(x-az)q = bx-ay. \checkmark \]
Result
The general solution is: