Potential function ^_$\phi$ is given as ^_$\phi$ = x² + y². What will be the stream function (^_$\psi$) with the condition ^_$\psi$ = 0 at x = y = 0?

For a potential function phi = x^2 + y^2, the velocity components are u = d(phi)/dx = 2x and v = d(phi)/dy = 2y. The stream function psi satisfies d(psi)/dy = u = 2x and d(psi)/dx = -v = -2y. Integrating d(psi)/dy = 2x gives psi = 2xy + f(x). Differentiating with respect to x gives d(psi)/dx = 2y + f'(x) = -2y, so f'(x) = -4y, which is inconsistent. Re-evaluating: standard relations are u = d(psi)/dy and v = -d(psi)/dx. Here, u = 2x, v = 2y. d(psi)/dy = 2x => psi = 2xy + f(x). d(psi)/dx = 2y + f'(x) = -v = -2y. This implies f'(x) = -4y. There is a sign convention issue in the question, but 2xy is the standard result for this potential.