We can use reduction of order. Given that [tex]y_1(x)=x[/tex] is a known solution, we look for a solution of the form [tex]y_2(x)=v(x)y_1(x)[/tex]. It has derivatives [tex]{y_2}'=v'y_1+v{y_1}'[/tex] and [tex]{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''[/tex]. Substituting these into the ODE gives
[tex]x(xv''+2v')-x(xv'+v)+xv=0[/tex]
[tex]x^2v''+(2x-x^2)v'=0[/tex]
Let [tex]w(x)=v'(x)[/tex] so that [tex]w'(x)=v''(x)[/tex] and we get an ODE linear in [tex]w[/tex]:
[tex]x^2w'+(2x-x^2)w=0[/tex]
Divide both sides by [tex]e^x[/tex]:
[tex]x^2e^{-x}w'+(2x-x^2)e^{-x}w=0/tex]
Since [tex](x^2e^{-x})=(2x-x^2)e^{-x}[/tex], we can condense the left side as the derivative of a product:
[tex](x^2e^{-x}w)'=0[/tex]
Integrate both sides and solve for [tex]w(x)[/tex]:
[tex]x^2e^{-x}w=C\implies w=\dfrac{Ce^x}{x^2}[/tex]
Integrate both sides again to solve for [tex]v(x)[/tex]. Unfortunately, there is no closed form for the integral of the right side, but we can leave the result in the form of a definite integral:
[tex]v=\displaystyle C_2+C_1\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]
where [tex]x_0[/tex] is any point on an interval over which a solution to the ODE exists.
Finally, multiply by [tex]y_1(x)[/tex] to solve for [tex]y_2(x)[/tex]:
[tex]y_2=\displaystyle C_2x+C_1x\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]
[tex]y_1(x)[/tex] already accounts for the [tex]C_2x[/tex] term above, so the second independent solution is
[tex]y_2=x\displaystyle\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]
