\(u = 4x = \frac{{\partial \psi }}{{\partial y}}\) 

\( \Rightarrow \partial \psi  = 4x\partial y\)

\( \Rightarrow \int {\partial \psi  = \int {4x\partial y} } \)

\( \Rightarrow \psi  = 4xy + f\left( x \right)\)

                                      \(v =  – 4y\) = -\(\frac{{ \partial \psi }}{{\partial x}}\)

\( \Rightarrow \frac{{\partial \psi }}{{\partial x}} = 4y\)

\( \Rightarrow \partial \psi  = 4y\partial x\)

\( \Rightarrow \int {\partial \psi  = \int {4y\partial x} } \)

\( \Rightarrow \psi  = 4xy + f\left( y \right)\)

On comparing these two equations for \(\psi \), stream functions is \(\psi  = 4xy + {\rm{constant}}\).

Also                               \(u = 4x = \frac{{\partial \phi }}{{\partial x}}\)

\( \Rightarrow 4x\partial x = \partial \phi \)

\( \Rightarrow \int {\partial \phi  = \int {4x\partial x} } \)

\( \Rightarrow \phi  = 4{x^2} + f\left( y \right)\)

                                      \(v =  – 4y = \frac{{\partial \phi }}{{\partial y}}\)

\( \Rightarrow \frac{{\partial \phi }}{{\partial y}} =  – 4y\)

\( \Rightarrow \int {\partial \phi  = \int { – 4y\partial y} } \)

\( \Rightarrow \phi  =  – 4{y^2} + f\left( x \right)\)

On comparing the above two equations, \(f\left( y \right) =  – 4{y^2}\) and \(f(x) = 4{x^2}\), velocity potential is \(\phi  = 4\left( {{x^2} – {y^2}} \right)\).

Now

\(\psi  = 4xy + {\rm{constant}}\), differentiating with respect to x, holding \(\psi  = {\rm{constant}}\)

\(0 = 4x\frac{{dy}}{{dx}} + 4y\)

\( \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\psi  = {\rm{constant}}}} =  – \frac{y}{x}\)

Differentiating \(\phi  = 4\left( {{x^2} – {y^2}} \right)\) with respect to x, holding \(\phi  = {\rm{constant}}\) 

\(0 = 2 \times 4x – 2 \times 4 \times y\frac{{dy}}{{dx}}\)

\( \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\phi  = {\rm{constant}}}} = \frac{x}{y}\)