http://eqworld.ipmnet.ru/en/solutions/ode/ode0202.pdf y http://eqworld.ipmnet.ru/en/solutions/ode/ode0207.pdf puede proporcionar sugerencias para usted.
Deje $s=x^n$ donde $n$ es una constante,
A continuación, $\dfrac{dy}{dx}=\dfrac{dy}{ds}\dfrac{ds}{dx}=nx^{n-1}\dfrac{dy}{ds}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(nx^{n-1}\dfrac{dy}{ds}\right)=nx^{n-1}\dfrac{d}{dx}\left(\dfrac{dy}{ds}\right)+n(n-1)x^{n-2}\dfrac{dy}{ds}=nx^{n-1}\dfrac{d}{ds}\left(\dfrac{dy}{ds}\right)\dfrac{ds}{dx}+n(n-1)x^{n-2}\dfrac{dy}{ds}=nx^{n-1}\dfrac{d^2y}{ds^2}nx^{n-1}+n(n-1)x^{n-2}\dfrac{dy}{ds}=n^2x^{2n-2}\dfrac{d^2y}{ds^2}+n(n-1)x^{n-2}\dfrac{dy}{ds}$
$\therefore n^2x^{2n-2}\dfrac{d^2y}{ds^2}+n(n-1)x^{n-2}\dfrac{dy}{ds}-xy=0$
$n^2x^{2n-3}\dfrac{d^2y}{ds^2}+n(n-1)x^{n-3}\dfrac{dy}{ds}-y=0$
$n^2s^{\frac{2n-3}{n}}\dfrac{d^2y}{ds^2}+n(n-1)s^{\frac{n-3}{n}}\dfrac{dy}{ds}-y=0$
La elección adecuada es $n=\frac{3}{2}$
$\therefore\frac{9}{4}\dfrac{d^2y}{ds^2}+\frac{3}{4}s^{-1}\dfrac{dy}{ds}-y=0$
$9s\dfrac{d^2y}{ds^2}+3\dfrac{dy}{ds}-4sy=0$
Deje $y=s^ku$ donde $k$ es una constante,
A continuación, $\dfrac{dy}{ds}=s^k\dfrac{du}{ds}+ks^{k-1}u$
$\dfrac{d^2y}{d^2s}=s^k\dfrac{d^2u}{ds^2}+ks^{k-1}\dfrac{du}{ds}+ks^{k-1}\dfrac{du}{ds}+k(k-1)s^{k-2}u=s^k\dfrac{d^2u}{ds^2}+2ks^{k-1}\dfrac{du}{ds}+k(k-1)s^{k-2}u$
$\therefore 9s\left(s^k\dfrac{d^2u}{ds^2}+2ks^{k-1}\dfrac{du}{ds}+k(k-1)s^{k-2}u\right)+3\left(s^k\dfrac{du}{ds}+ks^{k-1}u\right)-4ss^ku=0$
$9s^{k+1}\dfrac{d^2u}{ds^2}+18ks^k\dfrac{du}{ds}+9k(k-1)s^{k-1}u+3s^k\dfrac{du}{ds}+3ks^{k-1}u-4s^{k+1}u=0$
$9s^{k+1}\dfrac{d^2u}{ds^2}+(18k+3)s^k\dfrac{du}{ds}-(4s^{k+1}+3k(3k-2)s^{k-1})u=0$
$9s^2\dfrac{d^2u}{ds^2}+(18k+3)s\dfrac{du}{ds}-(4s^2+3k(3k-2))u=0$
La elección adecuada es $k=\frac{1}{3}$
$\therefore 9s^2\dfrac{d^2u}{ds^2}+9s\dfrac{du}{ds}-(4s^2+1)u=0$
Deje $t=ms$ donde $m$ es una constante,
A continuación, $\dfrac{du}{ds}=\dfrac{du}{dt}\dfrac{dt}{ds}=m\dfrac{du}{dt}$
$\dfrac{d^2u}{ds^2}=\dfrac{d}{ds}\left(m\dfrac{du}{dt}\right)=\dfrac{d}{dt}\left(m\dfrac{du}{dt}\right)\dfrac{dt}{ds}=m\dfrac{d^2u}{dt^2}m=m^2\dfrac{d^2u}{dt^2}$
$\therefore 9\left(\frac{t}{m}\right)^2m^2\dfrac{d^2u}{dt^2}+9\frac{t}{m}m\dfrac{du}{dt}-\left(4\left(\frac{t}{m}\right)^2+1\right)u=0$
$9t^2\dfrac{d^2u}{dt^2}+9t\dfrac{du}{dt}-\left(\frac{4t^2}{m^2}+1\right)u=0$
$t^2\dfrac{d^2u}{dt^2}+t\dfrac{du}{dt}-\left(\frac{4t^2}{9m^2}+\frac{1}{9}\right)u=0$
$t^2\dfrac{d^2u}{dt^2}+t\dfrac{du}{dt}+\left(\frac{4t^2}{9(mi)^2}-\frac{1}{9}\right)u=0$
La elección adecuada es $m=\frac{2i}{3}$
$\therefore t^2\dfrac{d^2u}{dt^2}+t\dfrac{du}{dt}+\left(t^2-\frac{1}{9}\right)u=0$