Substitute [tex]z=\ln x[/tex], so that
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}[/tex]
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)[/tex]
Then the ODE becomes
[tex]x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0[/tex]
[tex]\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0[/tex]
which has the characteristic equation [tex]r^2+1=0[/tex] with roots at [tex]r=\pm i[/tex]. This means the characteristic solution for [tex]y(z)[/tex] is
[tex]y_C(z)=C_1\cos z+C_2\sin z[/tex]
and in terms of [tex]y(x)[/tex], this is
[tex]y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)[/tex]
From the given initial conditions, we find
[tex]y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1[/tex]
[tex]y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8[/tex]
so the particular solution to the IVP is
[tex]y(x)=\cos(\ln x)+8\sin(\ln x)[/tex]
