 Solution

Use a parameter $t$. The characteristic equations areFor a graph of a function $z=f(x,y)$ satisfying the equation, solve the first two equations after substituting $z=f(x,y)$. The chain rule and the partial differential equation give $df(x(t),y(t))/dt=f+bx-ay$, exactly the third equation. Uniqueness for the resulting linear ordinary differential equation system shows that its characteristic through the initial point agrees with this curve on the graph of a function. This proves invariance as long as the graph of a function remains defined along the curve. A claim about the entire curve additionally requires a surface continued along that flow; a restricted graph of a function patch alone cannot imply global continuation.

Let $\mathbf{k}=(a,b,c)$, $\mathbf{\ell }=(l,m,n)$ and $\mathbf{r}=(x,y,z)$. Then $\dot{\mathbf{r}}=\mathbf{r}-\mathbf{k}\times \mathbf{r}$. Since $|\mathbf{k}|=1$,Along the characteristic starting from $s\mathbf{\ell }$ these give $\mathbf{k}\cdot \mathbf{r}=e^{t}s(\mathbf{k}\cdot \mathbf{\ell })$ and $|\mathbf{r}|=e^t|s|$. Consequently the characteristic surface generated by the given line satisfiesTo see that the whole displayed cone is generated, not just contained in it, decompose $\mathbf{\ell }=\kappa \mathbf{k}+\mathbf{v}$ with $\mathbf{v}\perp \mathbf{k}$, $\kappa =\mathbf{k}\cdot \mathbf{\ell }$. The full solution isWhen $0<|\kappa |<1$, variation of $t$ gives all azimuths about the axis, while the freely chosen $s$ supplies every positive radius on both halves. This is precisely the two-sided circular cone, withThe edge cases matter: $\kappa =0$ gives the plane $ax+by+cz=0$ (semi-angle $\pi /2$), and $|\kappa |=1$ generates only the axis (semi-angle zero), not a two-dimensional surface. In the latter case the initial line is itself characteristic, so it does not determine a unique integral surface. The nondegenerate cone is smooth away from its vertex and can be used as a graph of a function only on patches whose tangent plane is not vertical. Thus the printed geometric conclusion has its ordinary characteristic-surface meaning in the nondegenerate case, with these domain and degeneracy qualifications; it cannot be asserted for every global smooth graph of a function without them.