GATE Civil Engineering (CE) 2020 Shift 1 Solved Paper

We know that, ${\mathit{u}}_{\mathit{x}}=\frac{-\partial \mathit{\phi }}{\partial \mathit{x}}$
${\mathit{u}}_{\mathit{y}}=\frac{-\partial \mathit{\phi }}{\partial \mathit{y}}$
velocity in $\mathit{x}$ -direction, ${\mathit{u}}_{\mathit{x}}=\mathit{u}\sin \mathit{\theta }$
velocity in $\mathit{y}$ -direction, ${\mathit{u}}_{\mathit{y}}=\mathit{u}\cos \mathit{\theta }$
Now,
$\frac{-\partial \mathit{\phi }}{\partial \mathit{x}}=\mathit{u}\mathit{x}\Rightarrow \frac{-\partial \mathit{\phi }}{\partial \mathit{x}}=\mathit{u}\sin \mathit{\theta }$
Integrating above equation, we get
$\mathit{\varphi }=-\mathit{u}\sin \mathit{\theta }\mathit{x}+\mathit{f}\left(\mathit{y}\right)+\mathit{c}\to \left(1\right)$
Similarly
$\mathit{\varphi }=-\mathit{u}\cos \mathit{\theta }\mathit{y}+\mathit{f}\left(\mathit{x}\right)+\mathit{c}\to \left(2\right)$
From equation (1) and (2), we get
$\mathit{\varphi }=-\mathit{u}\left(\mathit{x}\sin \mathit{\theta }+\mathit{y}\cos \mathit{\theta }\right)$
when, $\frac{\partial \mathit{\phi }}{\partial \mathit{x}}={\mathit{u}}_{\mathit{x}}$ and $\frac{\partial \mathit{\phi }}{\partial \mathit{y}}={\mathit{u}}_{\mathit{y}}$
$\mathit{\varphi }=\mathit{u}\left(\mathit{x}\sin \mathit{\theta }+\mathit{y}\cos \mathit{\theta }\right)$
$\therefore \mathit{\varphi }=\pm \mathit{u}\left(\mathit{x}\sin \mathit{\theta }+\mathit{y}\cos \mathit{\theta }\right)$