Let [tex]y(x)=z(t)[/tex], where [tex]t=\ln x[/tex]. Then
[tex]\implies y'=\dfrac1xz'[/tex]
[tex]\implies y''=\dfrac1{x^2}(z''-z')[/tex]
so the ODE is equivalent to
[tex](z''-z')-4z'-6z=e^{3t}[/tex]
[tex]z''-5z'-6z=e^{3t}[/tex]
The characteristic equation is
[tex]r^2-5r-6=(r-6)(r+1)=0\implies r=6,r=-1[/tex]
so that the characteristic solution is
[tex]z_c=C_1e^{6t}+C_2e^{-t}[/tex]
For the particular solution, take
[tex]z_p=ae^{3t}[/tex]
[tex]\implies {z_p}'=3ae^{3t}[/tex]
[tex]\implies {z_p}''=9ae^{3t}[/tex]
[tex]\implies 9ae^{3t}-15ae^{3t}-6ae^{3t}=e^{3t}[/tex]
[tex]\implies -12a=1[/tex]
[tex]\implies a=-\dfrac1{12}[/tex]
[tex]\implies z_p=-\dfrac1{12}e^{3t}[/tex]
[tex]\implies y_p=-\dfrac1{12}x^3[/tex]
[tex]\implies y=y_c+y_p=C_1x^6+\dfrac{C_2}x-\dfrac1{12}x^3[/tex]
With the initial conditions, we get
[tex]y(1)=6\implies 6=C_1+C_2-\dfrac1{12}[/tex]
[tex]y'(1)=-1\implies -1=6C_1-C_2-\dfrac14[/tex]
[tex]\implies C_1=\dfrac{16}{21},C_2=\dfrac{149}{28}[/tex]
So the particular solution to the IVP is
[tex]y=\dfrac{16x^6}{21}+\dfrac{149}{28x}-\dfrac{x^3}{12}[/tex]
