Second order linear ODE question I am working on this equation: $ x^4y''+2x^3y'+y=0$ and i need a little help. Should I substitute y=exp(integrate(u)dx), and transform given equation into Riccati's: $ u'=-u^2-2u/x-1/x^4 $ (but i dont know know how to solve this either) or is there any other way? Thanks for tips and help
 A: It is interesting to note that the solution of 
$${x}^{4}{\frac {d^{2}}{d{x}^{2}}}y \left( x \right) +a{x}^{3}{\frac {d}
{dx}}y \left( x \right) +y \left( x \right) =0
$$
is given by
$$y \left( x \right) ={\it C_1}\,{x}^{{\frac {1}{2}}-{\frac {a}{2}}}
{{\it J}_{{\frac {1}{2}}-{\frac {a}{2}}}\left(\,{x}^{-1}\right)}+{\it C_2}\,{x}^{{\frac {1}{2}}-{\frac {a}{2}}}{{\it Y}_{{\frac {1}{2}}-{\frac {a}{2}}}\left(\,{x}^{-1}\right)}
$$
Then when $a=2$ we obtain 
$$y \left( x \right) ={\it C_1}\,\cos \left( {x}^{-1} \right) +{\it 
C_2}\,\sin \left( {x}^{-1} \right) 
$$
A: Let $t=\dfrac{1}{x}$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=-\dfrac{1}{x^2}\dfrac{dy}{dt}=-t^2\dfrac{dy}{dt}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-t^2\dfrac{dy}{dt}\right)=\dfrac{d}{dt}\left(-t^2\dfrac{dy}{dt}\right)\dfrac{dt}{dx}=\left(-t^2\dfrac{d^2y}{dt^2}-2t\dfrac{dy}{dt}\right)\left(-\dfrac{1}{x^2}\right)=\left(-t^2\dfrac{d^2y}{dt^2}-2t\dfrac{dy}{dt}\right)(-t^2)=t^4\dfrac{d^2y}{dt^2}+2t^3\dfrac{dy}{dt}$
$\therefore\dfrac{1}{t^4}\left(t^4\dfrac{d^2y}{dt^2}+2t^3\dfrac{dy}{dt}\right)-\dfrac{2}{t}\dfrac{dy}{dt}+y=0$
$\dfrac{d^2y}{dt^2}+\dfrac{2}{t}\dfrac{dy}{dt}-\dfrac{2}{t}\dfrac{dy}{dt}+y=0$
$\dfrac{d^2y}{dt^2}+y=0$
$y=C_1\sin t+C_2\cos t$
$y=C_1\sin\dfrac{1}{x}+C_2\cos\dfrac{1}{x}$
