Find the general solution of (∂ u)/(∂ x) + b (∂ u)/(∂ y) + c…
Find the general solution of \[ \frac{\partial u}{\partial x} + b\,\frac{\partial u}{\partial y} + c\,\frac{\partial u}{\partial z} = xyz, \] where $b$ and $c$ are real constants.
Explanation of Method
A Lagrange PDE with $P=1$, $Q=b$, $R=c$, $S=xyz$. The first two integrals are immediate; the third is found by expressing $y=c_1+bx$, $z=c_2+cx$ along a characteristic and integrating the resulting cubic polynomial $x(c_1+bx)(c_2+cx)$ directly with respect to $x$.
Step 1 · Characteristic equations
\[ \frac{dx}{1} = \frac{dy}{b} = \frac{dz}{c} = \frac{du}{xyz}. \tag{1} \]
Step 2 · First two integrals
From $dx=dy/b$: $y-bx=c_1$. \tag{I}
From $dx=dz/c$: $z-cx=c_2$. \tag{II}
Step 3 · Third integral
From $du=xyz\,dx$, substitute $y=c_1+bx$, $z=c_2+cx$ and expand: \[ du = \left[c_1c_2\,x+(c_1c+bc_2)x^2+bcx^3\right]dx. \] Integrating and collecting terms after substituting back $c_1=y-bx$, $c_2=z-cx$: \[ u = \frac{x^2yz}{2} - \frac{cx^3y}{6} - \frac{bx^3z}{6} + \frac{bcx^4}{12} + c_3, \] so \[ c_3 = u - \frac{x^2}{12}\bigl(6yz-2cxy-2bxz+bcx^2\bigr). \tag{III} \]
Step 4 · General solution
The general solution $c_3=\Phi(c_1,c_2)$ is: \[ u = \frac{x^2(6yz - 2cxy - 2bxz + bcx^2)}{12} + \Phi(y-bx,\;z-cx), \] where $\Phi$ is an arbitrary differentiable function of two arguments.
Verification
Let $f=\dfrac{x^2(6yz-2cxy-2bxz+bcx^2)}{12}$. Direct differentiation gives $f_y=\dfrac{x^2(6z-2cx)}{12}$, $f_z=\dfrac{x^2(6y-2bx)}{12}$, and (after expanding $f_x$ and combining) all cross terms in $f_x+bf_y+cf_z$ cancel except $xyz$: \[ f_x+bf_y+cf_z = \frac{12xyz}{12} = xyz. \checkmark \] For the homogeneous part, $\Phi_x+b\Phi_y+c\Phi_z = -b\Phi_1-c\Phi_2+b\Phi_1+c\Phi_2=0$ (with $\Phi_1,\Phi_2$ the partials of $\Phi$ in its two arguments). $\checkmark$ Hence $u_x+bu_y+cu_z=xyz$. $\checkmark$
Result
The general solution is: