I suspect there's a typo in the question, because [tex]y_1=x[/tex] is *not* a solution to the corresponding homogeneous equation. We have [tex]{y_1}'=1[/tex] and [tex]{y_1}''=0[/tex], so the ODE reduces to
[tex]0+7x+5x=12x\neq0[/tex]
Let [tex]y=x^m[/tex], then [tex]y'=mx^{m-1}[/tex] and [tex]y''=m(m-1)x^{m-2}[/tex], and substituting these into the (homogeneous) ODE gives
[tex]m(m-1)x^m+7mx^m+5x^m=0\implies m(m-1)+7m+5=m^2+6m+5=(m+5)(m+1)=0[/tex]
which then admits the characteristic solutions [tex]y_1=\dfrac1x[/tex] and [tex]y_2=\dfrac1{x^5}[/tex].
Now to find a solution to the non-homogeneous ODE. We look for a solution of the form [tex]y(x)=v(x)y_1(x)[/tex] or [tex]y(x)=v(x)y_2(x)[/tex].
It doesn't matter which one we start with, so let's use the first case. We get derivatives [tex]y'=x^{-1}v'-x^{-2}v[/tex] and [tex]y''=x^{-1}v''-2x^{-2}v'+2x^{-3}v[/tex]. Substituting into the ODE yields
[tex]x^2(x^{-1}v''-2x^{-2}v'+2x^{-3}v)+7x(x^{-1}v'-x^{-2}v)+5x^{-1}v=x[/tex]
[tex]xv''+5v'=x[/tex]
Substitute [tex]w=v'[/tex], so that [tex]w'=v''[/tex] and
[tex]xw'+5w=x[/tex]
which is linear in [tex]w[/tex], and we can condense the left side as the derivative of a product after multiplying both sides by [tex]x^4[/tex]:
[tex]x^5w'+5x^4=x^5\implies(x^5w)'=x^5\implies x^5w=\dfrac{x^6}6+C\implies w=\dfrac x6+\dfrac C{x^5}[/tex]
Integrate to solve for [tex]v[/tex]:
[tex]v=\dfrac{x^2}{12}+\dfrac{C_1}{x^4}+C_2[/tex]
Then multiply both sides by [tex]y_1=\dfrac1x[/tex] to solve for [tex]y[/tex]:
[tex]y=\dfrac x{12}+\dfrac{C_1}{x^5}+\dfrac{C_2}x[/tex]
so we found another fundamental solution [tex]y_3=x[/tex] that satisifes this ODE.
