Standard reduction of order procedure: suppose there is a second solution of the form [tex]y_2(x)=v(x)y_1(x)[/tex], which has derivatives
[tex]y_2=vx^4[/tex]
[tex]{y_2}'=v'x^4+4vx^3[/tex]
[tex]{y_2}''=v''x^4+8v'x^3+12vx^2[/tex]
Substitute these terms into the ODE:
[tex]x^2(v''x^4+8v'x^3+12vx^2)-7x(v'x^4+4vx^3)+16vx^4=0[/tex]
[tex]v''x^6+8v'x^5+12vx^4-7v'x^5-28vx^4+16vx^4=0[/tex]
[tex]v''x^6+v'x^5=0[/tex]
and replacing [tex]v'=w[/tex], we have an ODE linear in [tex]w[/tex]:
[tex]w'x^6+wx^5=0[/tex]
Divide both sides by [tex]x^5[/tex], giving
[tex]w'x+w=0[/tex]
and noting that the left hand side is a derivative of a product, namely
[tex]\dfrac{\mathrm d}{\mathrm dx}[wx]=0[/tex]
we can then integrate both sides to obtain
[tex]wx=C_1[/tex]
[tex]w=\dfrac{C_1}x[/tex]
Solve for [tex]v[/tex]:
[tex]v'=\dfrac{C_1}x[/tex]
[tex]v=C_1\ln|x|+C_2[/tex]
Now
[tex]y=C_1x^4\ln|x|+C_2x^4[/tex]
where the second term is already accounted for by [tex]y_1[/tex], which means [tex]y_2=x^4\ln x[/tex], and the above is the general solution for the ODE.
