i have a question of partial derivatives here:
given that the velocity components in two dimensional flow of a fluid, $$u=\frac{y^3}{3}+2x-x^2y \quad \text{&} \quad v=xy^2-2y-\frac{x^3}{3}$$ find out the stream line function $\psi(x, y)$
my trial:
for 2-D flow, $\frac{\partial \psi}{dy}=-u$ & $\frac{\partial \psi}{dx}=v$, plugging the values of u & v $$\frac{\partial \psi}{dy}=-\frac{y^3}{3}-2x+x^2y$$ integrating with respect to y, $$\psi=-\frac{y^4}{12}-2xy+\frac{x^2y^2}{2}+f(x) \ \ \ \ ..................\ \ (1)$$ differentiating $\psi$ with respect to x, $$\frac{\partial \psi}{dx}=-2y+xy^2+f'(x)$$ but $\frac{\partial \psi}{dx}=v=xy^2-2y-\frac{x^3}{3}$ now equating both, $$xy^2-2y-\frac{x^3}{3}=-2y+xy^2+f'(x)$$ $$f'(x)=-\frac{x^3}{3}$$ integrating with respect to x, $$f(x)=-\frac{x^4}{12}+C$$ C being constant of integration, plugging the value of f(x) in (1), stream line function $$\psi (x, y)=-\frac{y^4}{12}-2xy+\frac{x^2y^2}{2}-\frac{x^4}{12}+C$$$$=\frac{x^2y^2}{2}-\frac{x^4}{12}-\frac{y^4}{12}-2xy+C$$ but, my book says the answer is $\psi(x, y)=\frac{x^2y^2}{2}-\frac{x^4}{12}-\frac{y^4}{12}-2xy$, but i don't know the reason why constant of integration C is neglected.
somebody please help me explain or give me reasons. thank you