\(x\) and \(y\) components of velocity are given by \(u=cx\) and \(v=-cy\).We need to calculate the stream function and velocity potential.Here

\[u=cx
\\\frac{\partial \psi }{\partial y}=cx
\\\Rightarrow \partial \psi=cx\partial y
\\\Rightarrow \psi=cxy+f(x)
\\v=-cy=-\frac{\partial \psi}{\partial x}
\\\Rightarrow \psi=cxy+f(y)\]

Here f(x) and f(y) are constants:

\[\psi=cxy+const.
\\u=cx=\frac{d\phi}{dx}
\\\Rightarrow d\phi=cxdx
\\\Rightarrow \phi=cx^{2}+f(y)
\\v=-cy=\frac{\partial \phi}{\partial y}:\phi=-cy^{2}+f(x)\]

Therefore \[f(y)=-cy^{2}\;and \;f(x)=cx^{2}\]

and \[\phi=c\left ( x^{2}-y^{2} \right )
\\\psi=cxy+const.\]

Differentiating the above equation of \(\psi\) with respect to ‘\( x\)’, holding \(\psi=constant\)

\[\frac{\partial \psi}{\partial x}=cx\frac{dy}{dx}+cy\]

or \[\left ( \frac{dy}{dx} \right )_{\psi=constant}=\left ( \frac{-y}{x} \right )\]

Differentiating the above equation of \(\phi\) with respect to ‘\(x\)’,holding \(\phi=constant\)

\[0=2cx-2cy\left ( \frac{dy}{dx} \right )\]

or \[\left ( \frac{dy}{dx} \right )_{\phi=constant}=\left ( \frac{x}{y} \right )\]

On comparing the above equations we get \[\left ( \frac{dy}{dx} \right )_{\psi=constant}=\frac{-1}{\left (\frac{dy}{dx} \right )_{\phi=constant}}\]

Therefore the lines of constant \(\phi\) are perpendicular to the lines of constant \(\psi\).