$\partial_x^2[x^2p]+\partial_x[(x-1)p]=0$
$\partial_x[x^2p_x+2xp]+(x-1)p_x+p=0$
$x^2p_{xx}+2xp_x+2xp_x+2p+(x-1)p_x+p=0$
$x^2p_{xx}+(5x-1)p_x+3p=0$
$x^2\dfrac{d^2p}{dx^2}+(5x-1)\dfrac{dp}{dx}+3p=0$
This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0217.pdf
Let $\begin{cases}x=\dfrac{1}{u}\\p=u^3e^uy\end{cases}$ ,
Then $\dfrac{dp}{dx}=\dfrac{\dfrac{dp}{du}}{\dfrac{dx}{du}}=\dfrac{u^3e^u\dfrac{dy}{du}+(u^3+3u^2)e^uy}{-\dfrac{1}{u^2}}=-u^5e^u\dfrac{dy}{du}-(u^5+3u^4)e^uy$
$\dfrac{d^2p}{dx^2}=\dfrac{d}{dx}\left(-u^5e^u\dfrac{dy}{du}-(u^5+3u^4)e^uy\right)=\dfrac{\dfrac{d}{du}\left(-u^5e^u\dfrac{dy}{du}-(u^5+3u^4)e^uy\right)}{\dfrac{dx}{du}}=\dfrac{-u^5e^u\dfrac{d^2y}{du^2}-(2u^5+8u^4)e^u\dfrac{dy}{du}-(u^5+8u^4+12u^3)e^uy}{-\dfrac{1}{u^2}}=u^7e^u\dfrac{d^2y}{du^2}+2(u^7+3u^6)e^u\dfrac{dy}{du}+(u^7+8u^6+12u^5)e^uy$
$\therefore\dfrac{1}{u^2}\left(u^7e^u\dfrac{d^2y}{du^2}+(2u^7+8u^6)e^u\dfrac{dy}{du}+(u^7+8u^6+12u^5)e^uy\right)+\left(\dfrac{5}{u}-1\right)\left(-u^5e^u\dfrac{dy}{du}-(u^5+3u^4)e^uy\right)+3u^3e^uy=0$
$u^5e^u\dfrac{d^2y}{du^2}+(2u^5+8u^4)e^u\dfrac{dy}{du}+(u^5+8u^4+12u^3)e^uy-5u^4e^u\dfrac{dy}{du}-(5u^4+15u^3)e^uy+u^5e^u\dfrac{dy}{du}+(u^5+3u^4)e^uy+3u^3e^uy=0$
$u^5e^u\dfrac{d^2y}{du^2}+(3u^5+3u^4)e^u\dfrac{dy}{du}+(2u^5+6u^4)e^uy=0$
$u\dfrac{d^2y}{du^2}+(3u+3)\dfrac{dy}{du}+(2u+6)y=0$
Obviously, $y=e^{-2u}$ is a particular solution.
Let $y=e^{-2u}v$ ,
Then $\dfrac{dy}{du}=e^{-2u}\dfrac{dv}{du}-2e^{-2u}v$
$\dfrac{d^2y}{du^2}=e^{-2u}\dfrac{d^2v}{du^2}-4e^{-2u}\dfrac{dv}{du}+4e^{-2u}v$
$\therefore u\left(e^{-2u}\dfrac{d^2v}{du^2}-4e^{-2u}\dfrac{dv}{du}+4e^{-2u}v\right)+(3u+3)\left(e^{-2u}\dfrac{dv}{du}-2e^{-2u}v\right)+(2u+6)e^{-2u}v=0$
$u\left(\dfrac{d^2v}{du^2}-4\dfrac{dv}{du}+4v\right)+(3u+3)\left(\dfrac{dv}{du}-2v\right)+(2u+6)v=0$
$u\dfrac{d^2v}{du^2}-4u\dfrac{dv}{du}+4uv+(3u+3)\dfrac{dv}{du}-(6u+6)v+(2u+6)v=0$
$u\dfrac{d^2v}{du^2}-(u-3)\dfrac{dv}{du}=0$
$u\dfrac{d^2v}{du^2}=(u-3)\dfrac{dv}{du}$
$\dfrac{\dfrac{d^2v}{du^2}}{\dfrac{dv}{du}}=1-\dfrac{3}{u}$
$\int\dfrac{\dfrac{d^2v}{du^2}}{\dfrac{dv}{du}}du=\int\left(1-\dfrac{3}{u}\right)du$
$\ln\dfrac{dv}{du}=u-3\ln u+c_1$
$\dfrac{dv}{du}=\dfrac{c_2e^u}{u^3}$
$v=\int\dfrac{c_2e^u}{u^3}du$
$y=e^{-2u}\int\dfrac{c_2e^u}{u^3}du$
$y=e^{-\frac{2}{x}}\int c_2x^3e^{\frac{1}{x}}d\left(\dfrac{1}{x}\right)$
$y=e^{-\frac{2}{x}}\int C_2xe^{\frac{1}{x}}dx$
$y=e^{-\frac{2}{x}}(C_2\int_k^xxe^{\frac{1}{x}}dx+C_1)$
$y=C_1e^{-\frac{2}{x}}+C_2e^{-\frac{2}{x}}\int_k^xxe^{\frac{1}{x}}dx$
$p=\dfrac{C_1e^{-\frac{1}{x}}}{x^3}+\dfrac{C_2e^{-\frac{1}{x}}}{x^3}\int_k^xxe^{\frac{1}{x}}dx$