1. **Problem statement:** Given the function $u = f(x,y) = 2x^2 - 2y^2$, prove or verify it satisfies a partial differential equation (PDE). 2. **Step 1: Compute partial derivatives** - Compute $u_x = \frac{\partial u}{\partial x}$ and $u_y = \frac{\partial u}{\partial y}$. $$u_x = \frac{\partial}{\partial x}(2x^2 - 2y^2) = 4x$$ $$u_y = \frac{\partial}{\partial y}(2x^2 - 2y^2) = -4y$$ 3. **Step 2: Compute second partial derivatives** $$u_{xx} = \frac{\partial}{\partial x}(u_x) = \frac{\partial}{\partial x}(4x) = 4$$ $$u_{yy} = \frac{\partial}{\partial y}(u_y) = \frac{\partial}{\partial y}(-4y) = -4$$ 4. **Step 3: Verify the PDE** If the PDE is $u_{xx} + u_{yy} = 0$, then substitute: $$u_{xx} + u_{yy} = 4 + (-4) = 0$$ This shows $u$ satisfies the Laplace equation. 5. **Summary:** The function $u = 2x^2 - 2y^2$ satisfies the PDE $u_{xx} + u_{yy} = 0$ because the sum of its second partial derivatives equals zero.