The solution of $y_2-2y_1+y=0$ is

The given equation can be written as

$\left(\frac{\mathrm{d}}{\mathrm{d}x}-1\right)\left(\frac{\mathrm{d}y}{\mathrm{d}x}-y\right)=0$

If $\frac{\mathrm{d}y}{\mathrm{d}x}-y=u$ then (i) reduces to $\frac{\mathrm{d}u}{\mathrm{d}x}-u=0\Rightarrow u=$

$c_1{e}^x$. There fore, we have  $\frac{\mathrm{d}y}{\mathrm{d}x}-y=c_1{e}^x$ which is a linear 

equation whose I.F. is $e^{-x}.$ so $\frac{\mathrm{d}}{\mathrm{d}x}\left(y{e}^{-x}\right)=c_1\Rightarrow y{e}^{-x}=c_{1}x$

$+c_2\Rightarrow y=\left(c_{1}x+c_2\right)e^x$.(For a short cut see theory).