In fact it belongs to an Emden-Fowler equation.
First, according to http://eqworld.ipmnet.ru/en/solutions/ode/ode0302.pdf or http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=6 , all Emden-Fowler equations can be transformed into Abel equation of the second kind.
Let $\begin{cases}u=\dfrac{x^3}{y_n^\frac{3}{2}}\\v=\dfrac{x}{y_n}\dfrac{dy_n}{dx}\end{cases}$ ,
Then $\dfrac{dv}{du}=\dfrac{\dfrac{dv}{dx}}{\dfrac{du}{dx}}=\dfrac{\dfrac{x}{y_n}\dfrac{d^2y_n}{dx^2}+\dfrac{1}{y_n}\dfrac{dy_n}{dx}-\dfrac{x}{y_n^2}\left(\dfrac{dy_n}{dx}\right)^2}{\dfrac{3x^2}{y_n^\frac{3}{2}}-\dfrac{3x^3}{2y_n^\frac{5}{2}}\dfrac{dy_n}{dx}}=\dfrac{\dfrac{x}{y_n}\dfrac{d^2y_n}{dx^2}+\dfrac{v}{x}-\dfrac{v^2}{x}}{\dfrac{3u}{x}-\dfrac{3uv}{2x}}=\dfrac{\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}+v-v^2}{3u\left(1-\dfrac{v}{2}\right)}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}=\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}+v-v^2$
$\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}=3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v$
$\dfrac{d^2y_n}{dx^2}=\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)$
$\therefore\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)-\dfrac{nx}{\sqrt{y_n}}=0$
$\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)=\dfrac{nx}{\sqrt{y_n}}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v=\dfrac{nx^3}{y_n^\frac{3}{2}}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v=nu$
$3u\left(\dfrac{v}{2}-1\right)\dfrac{dv}{du}=v^2-v-nu$
Let $w=\dfrac{v}{2}-1$ ,
Then $v=2w+2$
$\dfrac{dv}{du}=2\dfrac{dw}{du}$
$\therefore6uw\dfrac{dw}{du}=(2w+2)^2-(2w+2)-nu$
$6uw\dfrac{dw}{du}=4w^2+6w+2-nu$
$w\dfrac{dw}{du}=\dfrac{2w^2}{3u}+\dfrac{w}{u}+\dfrac{2-nu}{6u}$
In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=\dfrac{1}{z}$ ,
Then $\dfrac{dw}{du}=-\dfrac{1}{z^2}\dfrac{dz}{du}$
$\therefore-\dfrac{1}{z^3}\dfrac{dz}{du}=\dfrac{2}{3uz^2}+\dfrac{1}{uz}+\dfrac{2-nu}{6u}$
$\dfrac{dz}{du}=\dfrac{(nu-2)z^3}{6u}-\dfrac{z^2}{u}-\dfrac{2z}{3u}$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2.
Hint: The equation $ty''-(t+1) y'+y=0$ can be rearranged as:
$$ty''-(t+1) y'+y=0$$
$$0=ty''-(t+1)y'+y=ty''-ty'-y'+y=t(y''-y')-(y'-y)\\
\Leftrightarrow t(y''-y')=(y'-y)\\
\Leftrightarrow t\dfrac{d}{dt}(y'-y)=(y'-y).$$
Substituting $u=y'-y$, this becomes the 1st order ODE $$tu'=u\\
\Leftrightarrow\dfrac{u'}{u}=\dfrac{1}{t}\\
\Leftrightarrow\int\dfrac{u'}{u}dt=\int\dfrac{du}{u}=\int\dfrac{1}{t}dt\\
\Leftrightarrow\log{|u|}=\log{|t|}+C\\
\Leftrightarrow u(t)=c_1t$$
Once you solve for $u(t)$, the problem reduces to solving the linear 1st order ODE with constant coefficients, $y'-y=u(t)=c_1 t$.
$$y'-y=u(t)=c_1 t\\
\implies e^{-t}y'-e^{-t}y=u(t)=c_1 t e^{-t}\\
\implies \dfrac{d}{dt}\left(e^{-t}y\right)=c_1 t e^{-t}\\
\implies e^{-t}y = c_2 + c_1\int t e^{-t} dt\\
\implies e^{-t}y = c_2 + c_1 e^{-t}(t+1)\\
\implies y(t) = c_2 e^{t} + c_1 (t+1)$$
Best Answer
By inspection, $y=x$ is a solution to the homogeneous equation, so you could try reduction of order to find the general solution to the homogeneous equation, and then variation of constants to find a particular solution matching the right-hand side.