1. **State the problem:** We are given the partial differential equation (PDE) $$\frac{\partial u}{\partial x} - 2 \frac{\partial^2 u}{\partial y^2} = 0$$ where $u = u(x,y)$ is an unknown function. We need to check which of the given functions $$U_1(x,y) = e^{2x - y}, \quad U_2(x,y) = \sin(2x - y), \quad U_3(x,y)$$ satisfy this PDE. 2. **Recall the PDE:** The PDE is $$\frac{\partial u}{\partial x} - 2 \frac{\partial^2 u}{\partial y^2} = 0$$ which means the partial derivative of $u$ with respect to $x$ minus twice the second partial derivative of $u$ with respect to $y$ must be zero. 3. **Check $U_1(x,y) = e^{2x - y}$:** - Compute $$\frac{\partial U_1}{\partial x} = \frac{\partial}{\partial x} e^{2x - y} = 2 e^{2x - y}$$ - Compute $$\frac{\partial U_1}{\partial y} = \frac{\partial}{\partial y} e^{2x - y} = - e^{2x - y}$$ - Compute $$\frac{\partial^2 U_1}{\partial y^2} = \frac{\partial}{\partial y} \left(- e^{2x - y}\right) = e^{2x - y}$$ - Substitute into PDE: $$2 e^{2x - y} - 2 \times e^{2x - y} = 2 e^{2x - y} - 2 e^{2x - y} = 0$$ Thus, $U_1$ satisfies the PDE. 4. **Check $U_2(x,y) = \sin(2x - y)$:** - Compute $$\frac{\partial U_2}{\partial x} = \frac{\partial}{\partial x} \sin(2x - y) = 2 \cos(2x - y)$$ - Compute $$\frac{\partial U_2}{\partial y} = \frac{\partial}{\partial y} \sin(2x - y) = - \cos(2x - y)$$ - Compute $$\frac{\partial^2 U_2}{\partial y^2} = \frac{\partial}{\partial y} \left(- \cos(2x - y)\right) = \sin(2x - y)$$ - Substitute into PDE: $$2 \cos(2x - y) - 2 \times \sin(2x - y) = 2 \cos(2x - y) - 2 \sin(2x - y) \neq 0$$ So $U_2$ does not satisfy the PDE. 5. **Check $U_3(x,y)$:** Since $U_3$ is not defined, we cannot verify it. **Final answer:** Only $$U_1(x,y) = e^{2x - y}$$ satisfies the PDE.