Here is my solution.

**Solution. **Let $\displaystyle p:=\frac{\mathrm{d}y}{\mathrm{d}x

}$, then by the chain rule we have $\displaystyle \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}

}=p\frac{\mathrm{d}p}{\mathrm{d}y}$.

Hence the given differential equation reduces to

$\displaystyle yp\mathrm{d}p-\big(2p^{2}+y^{2}\ln(y)\big)\mathrm{d}y=0.$

........(1)

Clearly, this equation is not exact since $\displaystyle \frac{\partial U}{\partial p}-\frac{\partial V}{\partial y}\neq0$, where $\displaystyle U(p,y):=yp$ and $\displaystyle V(p,y):=-\big(2p^{2}+y^{2}\ln(y)\big)$.

The integrating factor for this equation is

$\displaystyle \mu(y):=\exp\bigg\{-\int\frac{\frac{\partial U}{\partial p}-\frac{\partial V}{\partial y}}{U}\mathrm{d}y\bigg\}=\exp\bigg\{-5\int\frac{1}{y}\mathrm{d}y\bigg\}=\frac{1}{y^{5}} .$

Grouping the terms after multiplying through (1) by $\displaystyle \mu$, we get

$\displaystyle \frac{1}{2}\bigg(\frac{2py^{4}\mathrm{d}p-4p^{2}y^{3}\mathrm{d}y}{\big(y^{4}\big)^{2}}\bigg) +\frac{\ln(y)}{y^{3}}\mathrm{d}y=0$

or equivalently

$\displaystyle \frac{1}{2}\mathrm{d}\bigg(\frac{p^{2}}{y^{4}}\big g)+\frac{\ln(y)}{y^{3}}\mathrm{d}y=0.$

From this, we are lead to

$\displaystyle p=\pm\sqrt{c_{1}+\frac{y^{2}}{2}\big(1+2\ln(y)\big )},$

which gives us

$\displaystyle x=\pm\int\frac{1}{\sqrt{c_{1}+\frac{y^{2}}{2}\big( 1+2\ln(y)\big)}}\mathrm{d}y+c_{2}$

by using the relation $\displaystyle \mathrm{d}x=p\mathrm{d}y$.

This is the desired solution to (1).

....$\displaystyle \rule{0.3cm}{0.3cm}$

It seems not possible to solve the last integral.

I think there may be other solutions for this equation which may give explicit result.