Reduce the order of the ODE by setting [tex]z=y'[tex], so that [tex]z'=y''[/tex] and [tex]z''=y'''[/tex].
[tex]5z''-z'-6z=1+x^2[/tex]
Consider the homogeneous ODE
[tex]5z''-z'-6z=0[/tex]
which has characteristic equation
[tex]5r^2-r-6=(5r-6)(r+1)=0[/tex]
which has roots at [tex]r=\dfrac65[/tex] and [tex]r=-1[/tex], so that the characteristic solution is
[tex]z_c=C_1e^{6x/5}+C_2e^{-x}[/tex]
For the nonhomogeneous ODE,
[tex]5z''-z'-6z=1+x^2[/tex]
we can expect a particular solution of the form
[tex]z_p=ax^2+bx+c[/tex]
[tex]{z_p}'=2ax+b[/tex]
[tex]{z_p}''=2a[/tex]
Substituting these expressions into the ODE yields
[tex]10a-(2ax+b)-6(ax^2+bx+c)=1+x^2[/tex]
[tex]\iff-6ax^2+(-2a-b)x+(10a-b-6c)=x^2+1[/tex]
from which it follows that
[tex]\begin{cases}-6a=1\\-2a-b=0\\10a-b-6c=1\end{cases}\implies a=-\dfrac16,b=\dfrac13,c=-\dfrac12[/tex]
and so the particular solution is
[tex]z_p=-\dfrac16x^2+\dfrac13x-\dfrac12[/tex]
and the general solution for [tex]z[/tex] is
[tex]z=z_c+z_p[/tex]
[tex]z=C_1e^{6x/5}+C_2e^{-x}-\dfrac16x^2+\dfrac13x-\dfrac12[/tex]
Integrate both sides once to solve for [tex]y[/tex]:
[tex]z=y'\implies\displaystyle\int z\,\mathrm dx=\int y'\,\mathrm dx=y[/tex]
[tex]\implies y=\hat{C_1}e^{6x/5}+\hat{C_2}e^{-x}-\dfrac19x^3+\dfrac16x^2-\dfrac12x+\hat{C_3}[/tex]
