Find the general solution of \( y^{\prime \prime}-2 y^{\prime}+y=0 \) \[ \begin{array}{l}\text { A. } y(x)=c_{1} \mathbf{e}^{x}+c_{2} x \mathrm{e}^{x} \\ \text { B. } y(x)=c_{1} e^{x}+c_{2} e^{x} \\ \text { C. } y(x)=c_{1} x+c_{2} x \ln x \\ \text { D. } y(x)=c_{1} x+c_{2} x^{-1}\end{array} \]

To solve the differential equation \( y^{\prime \prime} - 2y^{\prime} + y = 0 \), we can start by finding the characteristic equation, which is obtained by assuming a solution of the form \( y = e^{rx} \). Substituting into the equation leads to \( r^2 - 2r + 1 = 0 \), which simplifies to \((r - 1)^2 = 0\). This means we have a repeated root \( r = 1 \). For a second-order linear differential equation with constant coefficients, if we have a repeated root, the general solution takes the form: \[ y(x) = c_1 e^{rx} + c_2 x e^{rx} \] Since our root is \( r = 1 \), we substitute it in to find: \[ y(x) = c_1 e^{x} + c_2 x e^{x} \] Therefore, the correct answer is: **A.** \( y(x) = c_{1} e^{x} + c_{2} x e^{x} \) Now, let's add a fun fact! In the world of differential equations, the method of characteristic equations is a bit like cracking a secret code. You start with a mysterious message (the differential equation) and decode it into a solvable format. It’s like being a mathematical detective! Another interesting tidbit is that repeated roots are akin to having a backup dancer in a performance. When one singer (solution) is strong, the backup (the additional \( x e^{x} \)) supports the performance, ensuring it still shines in the spotlight! This is a hallmark of repeated roots in the realm of differential equations.