(9) If \( u(x, y)=3 x^{2}+2 x-y^{3}-2 y^{2} \), find \( v(x, y) \) the harmonic conjugate of u (3mks)

To find the harmonic conjugate \( v(x, y) \) of the harmonic function \( u(x, y) = 3x^2 + 2x - y^3 - 2y^2 \), we first need to compute the partial derivatives of \( u \): 1. Calculate \( u_x \) and \( u_y \): \[ u_x = \frac{\partial u}{\partial x} = 6x + 2 \] \[ u_y = \frac{\partial u}{\partial y} = -3y^2 - 4y \] 2. From the Cauchy-Riemann equations, we know that \( v_x = u_y \) and \( v_y = -u_x \). Therefore: \[ v_x = -3y^2 - 4y \] \[ v_y = - (6x + 2) = -6x - 2 \] 3. To find \( v \), we will integrate \( v_x \) with respect to \( x \): \[ v(x, y) = \int (-3y^2 - 4y) \, dx = -3y^2x - 4xy + h(y) \] where \( h(y) \) is an arbitrary function of \( y \). 4. Next, we differentiate \( v \) with respect to \( y \) and compare it to \( v_y \): \[ v_y = \frac{\partial}{\partial y}(-3y^2x - 4xy + h(y)) = -6xy - 4x + h'(y) \] 5. Set this equal to \( -6x - 2 \): \[ -6xy - 4x + h'(y) = -6x - 2 \] 6. Simplifying gives \( h'(y) = -2 \). Integrating, we find \( h(y) = -2y + C \). 7. Therefore, we can substitute back into our equation for \( v \): \[ v(x, y) = -3y^2x - 4xy - 2y + C \] In conclusion, the harmonic conjugate \( v(x, y) \) of \( u(x, y) \) is: \[ v(x, y) = -3y^2x - 4xy - 2y + C \]