142. Find the linearly independent particular solutions and the general solution of the equation $y^{\prime \prime}-4 y^{\prime}+4 y$.
Solution. The characteristic equation $k^{2}-4 k+4=0$ has equal roots: $k_{1}=k_{2}=2$.

One particular solution is $y_{1}=e^{2 x}$. We will show that $y_{2}=x e^{2 x}$. We have $y^{\prime}=e^{2 x}+2 x e^{2 x}, y^{\prime}=2 e^{2 x}+2 e^{2 x}+4 x e^{2 x}=4 e^{2 x}+4 x e^{2 x}$, from which

$$
4 e^{2 x}+4 x e^{2 x}-4\left(e^{2 x}+2 x e^{2 x}\right)+4 x e^{2 x}=0
$$

i.e., we obtained an identity. Therefore, $y_{2}=x e^{2 x}$ is the second particular solution. These two solutions are linearly independent. Therefore, the general solution is written as

$$
y=C_{1} e^{2 x}+C_{2} x e^{2 x}=\left(C_{1}+C_{2} x\right) e^{2 x}
$$