 Solution

For $y=x$, the derivatives are $y^{\prime }=1$ and $y^{\prime \prime }=0$, and substitution gives $2(x-1)-2(x-1)=0$.

Use reduction of order: set $y=xv$, so $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$. Substitution and cancellation giveSince $x>0$, this reduces to $v^{\prime \prime }+2v^{\prime }=0$, hence $v^{\prime }=A{e}^{-2x}$ and $v=C_1+C_2{e}^{-2x}$. The general solution isThe two solutions are linearly independent, since their ratio $e^{-2x}$ is nonconstant; equivalently, their Wronskian is $-2x^2{e}^{-2x}\ne 0$.