1. **State the problem:** We are given the function $$\psi(x,y) = 3x^2 y + 2x^2 - y^3 - 2y^2$$. We need to show that $$\psi$$ is harmonic and then find its harmonic conjugate $$v(x,y)$$ using the Milne Thompson method. 2. **Check if $$\psi$$ is harmonic:** A function is harmonic if it satisfies Laplace's equation: $$\frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} = 0$$ Calculate the second partial derivatives: First, find $$\frac{\partial \psi}{\partial x}$$: $$\frac{\partial \psi}{\partial x} = \frac{\partial}{\partial x}(3x^2 y + 2x^2 - y^3 - 2y^2) = 6xy + 4x$$ Then, $$\frac{\partial^2 \psi}{\partial x^2}$$: $$\frac{\partial^2 \psi}{\partial x^2} = \frac{\partial}{\partial x}(6xy + 4x) = 6y + 4$$ Next, find $$\frac{\partial \psi}{\partial y}$$: $$\frac{\partial \psi}{\partial y} = \frac{\partial}{\partial y}(3x^2 y + 2x^2 - y^3 - 2y^2) = 3x^2 - 3y^2 - 4y$$ Then, $$\frac{\partial^2 \psi}{\partial y^2}$$: $$\frac{\partial^2 \psi}{\partial y^2} = \frac{\partial}{\partial y}(3x^2 - 3y^2 - 4y) = -6y - 4$$ Sum the second derivatives: $$\frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} = (6y + 4) + (-6y - 4) = 0$$ Since the sum is zero, $$\psi$$ is harmonic. 3. **Find the harmonic conjugate $$v(x,y)$$ using Milne Thompson method:** Recall that if $$\psi$$ is the imaginary part of an analytic function $$f(z) = u + iv$$, then $$u$$ and $$v$$ satisfy the Cauchy-Riemann equations: $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$ and $$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$. Since $$\psi$$ is given, we identify it as either $$u$$ or $$v$$. Here, $$\psi$$ is the imaginary part, so $$v = \psi$$ and $$u$$ is the harmonic conjugate we want to find. Using Milne Thompson method, the analytic function is: $$f(z) = u + iv = \phi + i\psi$$ where $$\phi$$ is the harmonic conjugate of $$\psi$$. The Milne Thompson formula for $$\phi$$ is: $$\phi(x,y) = \int \frac{\partial \psi}{\partial y} dx - \int \frac{\partial \psi}{\partial x} dy + C$$ Calculate $$\frac{\partial \psi}{\partial y}$$ and $$\frac{\partial \psi}{\partial x}$$ again: $$\frac{\partial \psi}{\partial y} = 3x^2 - 3y^2 - 4y$$ $$\frac{\partial \psi}{\partial x} = 6xy + 4x$$ Integrate $$\frac{\partial \psi}{\partial y}$$ with respect to $$x$$: $$\int (3x^2 - 3y^2 - 4y) dx = x^3 - 3y^2 x - 4y x + h(y)$$ Integrate $$-\frac{\partial \psi}{\partial x}$$ with respect to $$y$$: $$-\int (6xy + 4x) dy = -\int 6xy dy - \int 4x dy = -6x \frac{y^2}{2} - 4x y + k(x) = -3x y^2 - 4x y + k(x)$$ Equate the two expressions (ignoring arbitrary functions): $$x^3 - 3y^2 x - 4y x + h(y) = -3x y^2 - 4x y + k(x)$$ Simplify: $$x^3 + h(y) = k(x)$$ Since the left side depends on $$y$$ only through $$h(y)$$ and the right side depends on $$x$$ only through $$k(x)$$, both must be constant. Set constant zero for simplicity. Therefore, the harmonic conjugate is: $$\phi(x,y) = x^3 - 3x y^2 - 4x y$$ 4. **Final answer:** - $$\psi(x,y)$$ is harmonic. - The harmonic conjugate is: $$v(x,y) = \phi(x,y) = x^3 - 3x y^2 - 4x y$$