Every elementary flow in potential theory has its own signature form for $\varphi$. Match the given $\varphi = 5x - 12y$ against each of these signatures instead of differentiating first.
- Doublet: its potential is $\varphi = \dfrac{\kappa\cos\theta}{2\pi r}$, which blows up as $r \to 0$ and dies out far away. The given $\varphi$ is finite and well behaved everywhere, so it cannot be a doublet.
- Irrotational vortex: its potential is $\varphi = \dfrac{\Gamma}{2\pi}\theta$, a function of the polar angle $\theta$ only, and it is multi-valued (it keeps increasing as you go around the origin). Nothing like that shows up in $5x-12y$, so this is not a vortex.
- Source: its potential is $\varphi = \dfrac{\Lambda}{2\pi}\ln r$, a function of radial distance $r$, singular at the origin. The given $\varphi$ has no logarithm and no singularity, so it is not a source.
- Uniform flow: its potential is always a simple linear combination $\varphi = U_\infty x + V_\infty y$, where $U_\infty$ and $V_\infty$ are the constant free stream velocity components. Comparing term by term, $5x - 12y$ matches this pattern exactly, with $U_\infty = 5$ and $V_\infty = -12$.
Since the given potential has exactly the linear shape of a uniform stream, with constant coefficients and no dependence on $r$ or $\theta$, it must represent a uniform flow of speed $\sqrt{5^2+12^2}=13$ units, at a fixed direction.
Let's summarize:
- Sources and vortices involve $\ln r$ or $\theta$ and are singular at the origin.
- A doublet decays with $1/r$ and is also singular at the origin.
- A plain linear function of $x$ and $y$ is the signature of uniform flow.
So the correct option is (D), uniform flow.