146. Find the particular solution of the equation $y^{\prime \prime}-2 y^{\prime}+y=0$, if $y(0)=4, y^{\prime}(0)=2$.
Solution. The characteristic equation $k^{2}-2 k+1=0$ has equal real roots: $k_{2}=k_{1}=1$. Therefore, the general solution of the differential equation is the function $y=e^{x} \times$ $x\left(C_{1}+C_{2} x\right)$.

Next, substitute the initial conditions $y_{(0)}=4$ and $y_{(0)}^{\prime}=2$ into the equations $y=e^{x}\left(C_{1}+C_{2} x\right)$ and $y^{\prime}=e^{x}\left(C_{2} x+C_{1}+C_{2}\right)$. We obtain the system of equations $4=C_{1}, 2=C_{1}+C_{2}$, from which we determine $C_{2}=-2$.

Substituting the values $C_{1}=4, C_{2}=-2$ into the general solution $y=$ $=e^{x}\left(C_{1}+C_{2} x\right)$, we obtain the particular solution $y=e^{x}(4-2 x)$.