Solving a second order non-linear ODE - what's the deal with this one? Consider the following ODE:
$$3x^2y''^2 - 2(3xy' +y)y'' +4y'^2=0$$
So my first instinct would be to perform the following substitution: $y' = p$, $p = p(x)$, where $y'' = p'$
Which would transform my equation into the following:
$$3x^2p'^2 - 2(3xp + y)p' +p^2=0$$
However, this is a new one, because every time I did these kind of equations, I could divide the whole equation by a power of $y$ to get an equation consisting of only the variables $p$ and $x$ which could be solved. However, I can't do that here. Does anyone have any lead on how to solve this?
 A: This is linearizable by differentiation ode.
After differentiation we get ode
$$2y'''(3x^2y''-3xy'-y)=0.$$
I get final solutions
$$y=\frac{C_2^2x^2}{C_1}+C_2x+C_1,$$
$$y=C\, {{x}^{\frac{2}{\sqrt{3}}+1}},$$
$$y=C\, {{x}^{1-\frac{2}{\sqrt{3}}}}$$
A: As you have seen, applied like that this substitution only increases the number of variables. You get a homogeneous pattern similar to an Euler-Cauchy equation by multiplying the original equation by $x^2$,
\begin{align}
0&=3(x^2y'')^2-2(3xy'+y)(x^2y'')+4(xy')^2
\end{align}
One can now make this autonomous by borrowing the substitution $u(t)=y(e^t)$ from the Cauchy-Euler equation, $u'(t)=e^ty'(e^t)=xy'(x)$, $u''(t)=e^{2t}y''(t)+e^ty'(e^t)=x^2y''(x)+u'(t)$\begin{align}
0&=3(u''(t)-u'(t))^2-2(3u'(t)+u(t))(u''(t)-u'(t))+4u'(t)^2
\\
&=3u''^2-6u''u'+3u'^2-6u''u'+6u'^2-2u''u+2u'u+4u'^2
\\
&=3u''^2-12u''u'+13u'^2-2u''u+2u'u
\end{align}
As this is autonomous, one could insert $u'=v(u)$, $u''=v'(u)v(u)$. But even that does not appear helpful.
A: The expression can be rewritten as:
$$3x^2(y'')^2 - 6 x y'y'' - 2yy'' + 4(y')^2 = 0.$$
We make an observation as follows. Suppose that $y'' \neq 0$ for all $x$ in the appropriate domain. The above equation simplifies to:
$$3x^2 - 6 x \frac{y'}{y''} - 2\frac{y}{y''} + 4\left(\frac{y'}{y''}\right)^2 = 0.$$
Note that if we have:
$$\frac{y'}{y''} \sim x, \frac{y}{y''} \sim x^2, \frac{y'}{y''} \sim x.$$
This implies that we can guess a possible solution could take the form of a simple polynomial $y(x) = C_0 + C_1 x + C_2 x^2 + C_3 x^3 \ldots$ (This is due to the fact that the derivative of a polynomial is always one degree lower than that of the original polynomial.)
Recall that we have guessed that $y''\neq 0$ for all $x$. This thus inspires that we guess $y'' = D$ for some constant $D$. The equivalent "guess" for the solution would be
$$ y = A + Bx + Cx^2.$$
With that, substitute our guess into the main equation to obtain:
$$ 3x^2(2C)^2 - 6 x(B+2Cx)(2C) = 2(A + Bx + Cx^2)(2C) - 4(B+2Cx)^2.$$
Surprisingly, we obtain that the coefficient of $x^2$ and $x$ to be $0$. The equation is thus reduced to
$$B^2 = AC.$$
Recall that we are free to choose the values of $A,B$ and $C$, as long as they satisfy the constraint that we have just derived. In particular, if you have tried solving the equation using WolframAlpha, the suggested solution is given by:
$$y[x] \rightarrow x c_1 + x^2 \frac{c_1^2}{c_2} + c_2.$$
Since $A = c_2, B = c_1, C = \frac{c_1^2}{c_2}$, we indeed have $B^2 = AC$. The parameterization of the solution from WolframAlpha is just one particular parameterization for $B^2 = AC$.
