<s>[INST] Solve $x^4y^{\prime\prime} = (y-xy^\prime)^3, y(1) = y^\prime(1) = 1$ $\lambda^4x^4\lambda^{n-2}y^{\prime\prime} = (\lambda^ny-\lambda x\lambda^{n-1}y^\prime)^3 \Rightarrow \lambda^{n+2}x^4y^{\prime\prime} = (\lambda^n(y-xy^\prime))^3$. 
$\lambda^{n+2} = \lambda^{3n} \Rightarrow n+2 = 3n \Rightarrow n = 1$.
Let $x = e^t, y = ue^{nt} = ue^t$.
Now, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{u^\prime e^t + ue^t}{e^t} = u^\prime + u$. And, $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}}{\frac{dx}{dt}}(\frac{dy}{dx}) = \frac{\frac{d}{dt}(u^\prime + u)}{e^t} = e^{-t}(\frac{d^2u}{dt^2} + \frac{du}{dt})$.
Thus,
$e^{4t}(e^{-t}(\frac{du^2}{dt^2} + \frac{du}{dt})) = (ue^{t}-e^t(u+\frac{du}{dt}))^3 \Rightarrow e^{3t}(\frac{d^2u}{t^2} + \frac{du}{dt}) = e^{3t}(u-(u+\frac{du}{dt}))^3 \Rightarrow (\frac{d^2u}{dt^2} + \frac{du}{dt}) = (\frac{du}{dt})^3$.
Let $p = \frac{du}{dt}, p^\prime = \frac{d^2u}{dt^2} = \frac{dp}{du}\frac{du}{dt} = p\frac{dp}{du}$.
Thus,
$(p + p\frac{dp}{du}) = p^3 \Rightarrow 1 + \frac{dp}{du} = p^2 \Rightarrow \frac{dp}{du} = p^2-1 \Rightarrow u + c_1 = \int\frac{dp}{p^2-1} = \int\frac{dp}{(p-1)(p+1)} = \int(\frac{1}{2(p-1)} - \frac{1}{2(p+1)})dp = \frac{1}{2}\ln(p-1)-\frac{1}{2}\ln(p+1) = \ln(\frac{\sqrt{p-1}}{\sqrt{p+1}}) \Rightarrow c_1e^u = \frac{\sqrt{p-1}}{\sqrt{p+1}} \Rightarrow c_1e^{2u} = \frac{p-1}{p+1} \Rightarrow p-1 = c_1e^{2u}p + c_1e^{2u} \Rightarrow p = \frac{c_1e^{2u}+1}{1-c_1e^{2u}} = \frac{du}{dt}$.
But then I think that I have to use the initial conditions to get $c_1$, but I don know how to do that.
 [/INST] Taking $y = u x$, the equation becomes 
$$ u'' = - \dfrac{u' ((u')^2 x^2 + 2)}{x} $$
with $u(1) = 1$, $u'(1) = 0$.
If $u' = v$, this is
$$ v' = - \dfrac{v (v^2 x^2 + 2)}{x},\ v(1) = 0$$
Now we could go on to solve this differential equation in general, but for this specific initial condition it's obvious that the solution is $v = 0$.  Then $u$ must be a constant, namely $u=1$ from the initial condition, so $y=x$.
</s>