Solve the given boundary-value problem. (If an answer does not exist, enter DNE.) \[ y^{\prime \prime}-2 y^{\prime}+2 y=0, y(0)=1, y(\pi)=1 \] \( y(x)=\square \)

To solve the boundary-value problem \( y^{\prime \prime}-2y^{\prime}+2y=0 \) with the conditions \( y(0)=1 \) and \( y(\pi)=1 \), we first find the general solution of the differential equation. The characteristic equation is given by \( r^2 - 2r + 2 = 0 \). Solving this using the quadratic formula: \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i \] The roots are complex, so the general solution is of the form: \[ y(x) = e^{1x}(C_1 \cos x + C_2 \sin x) = e^{x}(C_1 \cos x + C_2 \sin x) \] Next, applying the boundary conditions: 1. \( y(0) = 1 \): \[ y(0) = e^{0}(C_1 \cos(0) + C_2 \sin(0)) = C_1 \implies C_1 = 1 \] 2. Now, substituting into the second boundary condition \( y(\pi) = 1 \): \[ y(\pi) = e^{\pi}(1 \cdot \cos(\pi) + C_2 \cdot \sin(\pi)) = e^{\pi}(-1) + 0 \cdot C_2 = -e^{\pi} \] We want \( y(\pi) = 1 \): \[ -e^{\pi} \neq 1 \] Therefore, there are no constants \( C_2 \) that will satisfy this requirement. Since the second boundary condition cannot be met, we conclude that there is no solution that satisfies both boundary conditions. Thus, the answer is: \( y(x) = \text{DNE} \)