Substitute [tex]x=e^t[/tex]. By the chain rule,
[tex]\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx}[/tex]
Now [tex]t=\ln(x)\implies\frac{dt}{dx}=\frac1x[/tex], so
[tex]\dfrac{dy}{dx} = \dfrac1x \dfrac{dy}{dt} \\\\ ~~~~~~~~ \iff \dfrac{dy}{dt} = x\dfrac{dy}{dx}[/tex]
Differentiate both sides again to recover the second and third derivatives.
[tex]\dfrac{d^2y}{dx^2} = -\dfrac1{x^2}\dfrac{dy}{dt} + \dfrac1x \dfrac{d\frac{dy}{dt}}{dx} \\\\ ~~~~~~~~ = -\dfrac1{x^2}\dfrac{dy}{dt} + \dfrac1x \left(\dfrac{d\frac{dy}{dt}}{dt}\cdot\dfrac{dt}{dx}\right) \\\\ ~~~~~~~~ = -\dfrac1{x^2} \dfrac{dy}{dt} + \dfrac1{x^2}\dfrac{d^2y}{dt^2} \\\\ ~~~~~~~~ \iff \dfrac{d^2y}{dt^2} - \dfrac{dy}{dt} = x^2 \dfrac{d^2y}{dx^2}[/tex]
[tex]\dfrac{d^3y}{dx^3} = \dfrac2{x^3}\dfrac{dy}{dt} - \dfrac1{x^2}\dfrac{d\frac{dy}{dt}}{dx} - \dfrac2{x^3} \dfrac{d^2y}{dt^2} + \dfrac1{x^2}\dfrac{d\frac{d^2y}{dt^2}}{dx} \\\\ ~~~~~~~~ = \dfrac2{x^3} \dfrac{dy}{dt} - \dfrac1{x^2}\left(\dfrac{d\frac{dy}{dt}}{dt}\cdot\dfrac{dt}{dx}\right) - \dfrac2{x^3} \dfrac{d^2y}{dt^2} + \dfrac1{x^2}\left(\dfrac{d\frac{d^2y}{dt^2}}{dt}\cdot\dfrac{dt}{dx}\right) \\\\ ~~~~~~~~ = \dfrac2{x^3} \dfrac{dy}{dt} - \dfrac1{x^3} \dfrac{d^2y}{dt^2} - \dfrac2{x^3} \dfrac{d^2y}{dt^2} + \dfrac1{x^3}\dfrac{d^3y}{dt^3} \\\\ ~~~~~~~~ \iff \dfrac{d^3y}{dt^2} - 3 \dfrac{d^2y}{dt^2} + 2 \dfrac{dy}{dt} = x^3 \dfrac{d^3y}{dx^3}[/tex]
The ODE then transforms to a linear one,
[tex]3\left(\dfrac{d^3y}{dt^2} - 3 \dfrac{d^2y}{dt^2} + 2 \dfrac{dy}{dt}\right) + 28\left(\dfrac{d^2y}{dt^2} - \dfrac{dy}{dt}\right) + 55\dfrac{dy}{dt} + 9 y = 0[/tex]
or using Lagrange's prime notation,
[tex]3(y''' - 3y'' + 2y') + 28(y'' - y') + 55y' + 9y = 0[/tex]
[tex]3y''' + 19y'' + 33y' + 9y = 0[/tex]
The characteristic equation is
[tex]3r^3 + 19r^2 + 33r + 9r = (r + 3)^2 \left(r + \dfrac13\right) = 0[/tex]
with roots at [tex]r=-3[/tex] and [tex]r=-\frac13[/tex], so the general solution is
[tex]y(t) = C_1 e^{-1/3\,t} + C_2 e^{-3t} + C_3 t e^{-3t}[/tex]
Back in terms of [tex]x[/tex], we get
[tex]\boxed{y(x) = C_1 x^{-1/3} + C_2 x^{-3} + C_3 x^{-3}\ln(x)}[/tex]
